Free math worksheets

- Each coordinate in the angle brackets is one component function. - Keep the components in the order in which they appear. - No differentiation or evaluation is needed.

1. Read each coordinate of the vector-valued function as a separate scalar function of \(t\). 2. The components are \(x(t)=3t-1\), \(y(t)=\sqrt{t+2}\), and \(z(t)=e^t\).

\(x(t)=3t-1\), \(y(t)=\sqrt{t+2}\), \(z(t)=e^t\)

- A vector-valued function is evaluated component by component. - Use the same parameter value in every coordinate. - Preserve the coordinate order in the final vector.

1. Substitute \(t=-1\) into each component. 2. This gives \(\mathbf r(-1)=\langle (-1)^2,2(-1)+1,3-(-1)\rangle=\langle1,-1,4\rangle\).

\(\langle1,-1,4\rangle\)

- A vector-valued function is defined only when every component is defined at the same parameter value. - Determine the allowed interval for each non-polynomial component separately. - Intersect the two sets rather than choosing one component's domain.

1. The square-root component requires \(t+2\ge0\), so \(t\ge-2\). 2. The logarithmic component requires \(3-t>0\), so \(t<3\). 3. The vector function is defined only where both conditions hold, giving \([-2,3)\).

\([-2,3)\)

- A two-dimensional position vector packages the x-coordinate and y-coordinate into one expression. - Match components to coordinates in order. - Keep the same parameter in both equations.

1. Match the first component with \(x(t)\) and the second component with \(y(t)\). 2. The parametric equations are \(x=2t+1\) and \(y=t^2-4\).

\(x=2t+1\), \(y=t^2-4\)

- A two-dimensional vector-valued function packages the two parametric coordinates together. - Preserve the coordinate order. - Use the same parameter that appears in the given equations.

1. Place the x-coordinate first and the y-coordinate second in the position vector. 2. Therefore, \(\mathbf r(t)=\langle1-2t,3t+4\rangle\).

\(\mathbf r(t)=\langle1-2t,3t+4\rangle\)

- A point on a vector-valued curve requires one parameter value to satisfy all coordinate equations simultaneously. - One coordinate may leave more than one candidate. - Compare the candidate sets rather than stopping after the first matching coordinate.

1. The first coordinate condition \(t^2=1\) gives candidates \(t=1\) and \(t=-1\). 2. The second coordinate condition \(t^2+t=2\) gives candidates \(t=1\) and \(t=-2\). 3. The same parameter must satisfy both component equations, so the only valid value is \(t=1\).

\(t=1\)

- Differentiate the polynomial in each vector component independently. - Preserve the original first-component/second-component order. - Substitute \(t=2\) only after writing the complete derivative vector.

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle 2t, 3t^{2}+2\right\rangle\). 2. Substitution gives \(\mathbf{r}'(2)=\left\langle 4, 14\right\rangle\).

\(\mathbf{r}'(t)=\left\langle 2t, 3t^{2}+2\right\rangle\); \(\mathbf{r}'(2)=\left\langle 4, 14\right\rangle\)

- Differentiate sine and cosine componentwise, including the negative sign from cosine. - Keep the derivative components in the same order as the original vector. - Use exact special-angle values at \(t=\frac{\pi}{4}\).

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle \cos(t), -\sin(t)\right\rangle\). 2. Substitution gives \(\mathbf{r}'\left(\frac{\pi}{4}\right)=\left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right\rangle\).

\(\mathbf{r}'(t)=\left\langle \cos(t), -\sin(t)\right\rangle\); \(\mathbf{r}'\left(\frac{\pi}{4}\right)=\left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right\rangle\)

- Differentiate the first exponential component directly. - Use the product rule on the second component \(te^{t}\). - Evaluate both derivative components at \(t=0\) after simplifying.

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle e^{t}, (t+1)e^{t}\right\rangle\). 2. Substitution gives \(\mathbf{r}'(0)=\left\langle 1, 1\right\rangle\).

\(\mathbf{r}'(t)=\left\langle e^{t}, (t+1)e^{t}\right\rangle\); \(\mathbf{r}'(0)=\left\langle 1, 1\right\rangle\)

- Respect the positive domain of the logarithmic component. - Rewrite the reciprocal component as a negative power before differentiating. - Evaluate the ordered derivative vector at \(t=1\).

1. On the domain \(t>0\), differentiate each component: \(\mathbf{r}'(t)=\left\langle \frac{1}{t},-\frac{1}{t^{2}}\right\rangle\). 2. Substitution gives \(\mathbf{r}'(1)=\left\langle 1,-1\right\rangle\).

\(\mathbf{r}'(t)=\left\langle \frac{1}{t},-\frac{1}{t^{2}}\right\rangle\) for \(t>0\); \(\mathbf{r}'(1)=\left\langle 1,-1\right\rangle\)

- Differentiate the cubic component with the power rule. - Rewrite the square root as a fractional power before differentiating. - Substitute \(t=4\) into each derivative component separately.

1. For \(t>0\), differentiate each component: \(\mathbf{r}'(t)=\left\langle 3t^{2},\frac{1}{2\sqrt{t}}\right\rangle\). The original vector function is defined at \(t=0\), but its square-root component is not differentiable there. 2. Substitution gives \(\mathbf{r}'(4)=\left\langle 48,\frac{1}{4}\right\rangle\).

\(\mathbf{r}'(t)=\left\langle 3t^{2},\frac{1}{2\sqrt{t}}\right\rangle\) for \(t>0\); \(\mathbf{r}'(4)=\left\langle 48,\frac{1}{4}\right\rangle\)

- Apply the chain rule separately to the double-angle and triple-angle components. - Keep the inner-rate factors attached to their corresponding components. - Use exact trigonometric values after substituting \(t=\frac{\pi}{6}\).

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle -2\sin(2t), 3\cos(3t)\right\rangle\). 2. Substitution gives \(\mathbf{r}'\left(\frac{\pi}{6}\right)=\left\langle -\sqrt{3}, 0\right\rangle\).

\(\mathbf{r}'(t)=\left\langle -2\sin(2t), 3\cos(3t)\right\rangle\); \(\mathbf{r}'\left(\frac{\pi}{6}\right)=\left\langle -\sqrt{3}, 0\right\rangle\)

- Differentiate the reciprocal term in each component with careful signs. - Preserve the plus/minus distinction between the two components. - Evaluate the fractions exactly at \(t=2\).

1. For \(t\ne0\), differentiate each component: \(\mathbf{r}'(t)=\left\langle 1-\frac{1}{t^{2}},1+\frac{1}{t^{2}}\right\rangle\). 2. Substitution gives \(\mathbf{r}'(2)=\left\langle \frac{3}{4},\frac{5}{4}\right\rangle\).

\(\mathbf{r}'(t)=\left\langle 1-\frac{1}{t^{2}},1+\frac{1}{t^{2}}\right\rangle\) for \(t\ne0\); \(\mathbf{r}'(2)=\left\langle \frac{3}{4},\frac{5}{4}\right\rangle\)

- Use the product rule on \(t^{2}e^{t}\) and factor the result if helpful. - Apply the chain rule to the negative exponent in the second component. - Substitute \(t=1\) only after both derivative components are complete.

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle t(t+2)e^{t}, -e^{-t}\right\rangle\). 2. Substitution gives \(\mathbf{r}'(1)=\left\langle 3e, -\frac{1}{e}\right\rangle\).

\(\mathbf{r}'(t)=\left\langle t(t+2)e^{t}, -e^{-t}\right\rangle\); \(\mathbf{r}'(1)=\left\langle 3e, -\frac{1}{e}\right\rangle\)

- Compare each component of the student's derivative with the corresponding original component. - A vector derivative should reflect a derivative operation in every component. - Preserve the original component order when writing the corrected vector.

1. A vector-valued function is differentiated one component at a time. 2. The student did not differentiate the second component. Since \(\frac{d}{dt}\sin(t)=\cos(t)\), the correct derivative is \(\mathbf{r}'(t)=\left\langle 2t, \cos(t)\right\rangle\).

\(\mathbf{r}'(t)=\left\langle 2t, \cos(t)\right\rangle\)

- The two proposals differ only in the derivative of the second component, so focus your check there. - Ask whether \(te^t\) is a single basic function or a product of two changing factors. - After differentiating that component independently, compare it with the two choices.

1. Differentiate each component separately. 2. The second component requires the product rule: \(\frac{d}{dt}(te^{t})=e^{t}+te^{t}=(t+1)e^{t}\). 3. Therefore, \(\mathbf{r}'(t)=\left\langle e^{t}, (t+1)e^{t}\right\rangle\).

\(\mathbf{r}'(t)=\left\langle e^{t}, (t+1)e^{t}\right\rangle\); the other proposal omits part of the product-rule derivative.

- Apply the chain rule to the double-angle cosine component. - Differentiate the sine component independently. - Retain the inner factor from \(2t\) in the first component.

1. Differentiate each component separately. 2. The chain rule gives \(\frac{d}{dt}\cos(2t)=-2\sin(2t)\). 3. Therefore, \(\mathbf{r}'(t)=\left\langle -2\sin(2t), \cos(t)\right\rangle\).

\(\mathbf{r}'(t)=\left\langle -2\sin(2t), \cos(t)\right\rangle\)

- A vector’s components are differentiated separately, not multiplied together. - Use the logarithm rule for the first component and the power rule for the square root. - Preserve the domain restriction \(t>0\) in the corrected result.

1. The angle brackets list separate components; they do not indicate multiplication. 2. Differentiate each component separately to obtain \(\mathbf{r}'(t)=\left\langle \frac{1}{t}, \frac{1}{2\sqrt{t}}\right\rangle\) for \(t>0\).

The components are not factors of a product. The derivative is \(\mathbf{r}'(t)=\left\langle \frac{1}{t}, \frac{1}{2\sqrt{t}}\right\rangle\) for \(t>0\).

- Derive one tangent-vector formula before substituting either value. - Evaluate both components at \(t=0\) and \(t=1\). - Compare magnitude and direction, including whether either vector is zero.

1. Differentiate componentwise: \(\mathbf{r}'(t)=\left\langle 2t, 3t^{2}\right\rangle\). 2. At \(t=0\), the derivative is \(\left\langle 0, 0\right\rangle\). 3. At \(t=1\), the derivative is \(\left\langle 2, 3\right\rangle\). 4. The curve is momentarily stationary at \(t=0\), while at \(t=1\) it has a nonzero tangent vector pointing right and upward.

At \(t=0\): \(\left\langle 0, 0\right\rangle\), so the curve is momentarily stationary. At \(t=1\): \(\left\langle 2, 3\right\rangle\), a nonzero vector pointing right and upward.

- Differentiate the polynomial components once and use the same formula at both values. - Substitute the negative parameter carefully into even and odd powers. - Compare both vector magnitudes and signs of corresponding components.

1. Differentiate componentwise: \(\mathbf{r}'(t)=\left\langle 3t^{2}-3, 2t\right\rangle\). 2. At \(t=-1\), the derivative is \(\left\langle 0, -2\right\rangle\). 3. At \(t=1\), the derivative is \(\left\langle 0, 2\right\rangle\). 4. The vectors have the same magnitude but point in opposite vertical directions.

At \(t=-1\): \(\left\langle 0, -2\right\rangle\) At \(t=1\): \(\left\langle 0, 2\right\rangle\) The vectors have equal magnitude and opposite directions.

- Differentiate the circular parameterization componentwise. - Evaluate exact trigonometric values at both parameters. - Compare lengths and use a dot product to check the angle between the vectors.

1. Differentiate componentwise: \(\mathbf{r}'(t)=\left\langle -\sin(t), \cos(t)\right\rangle\). 2. At \(t=0\), the derivative is \(\left\langle 0, 1\right\rangle\). 3. At \(t=\frac{\pi}{2}\), the derivative is \(\left\langle -1, 0\right\rangle\). 4. The vectors are perpendicular and have the same magnitude, \(1\).

At \(t=0\): \(\left\langle 0, 1\right\rangle\) At \(t=\frac{\pi}{2}\): \(\left\langle -1, 0\right\rangle\) The vectors are perpendicular and have equal magnitude.

- Use the product rule to obtain one derivative-vector formula. - Evaluate it exactly at \(t=0\) and \(t=1\). - Compare component signs, slopes, and magnitudes rather than only one coordinate.

1. Differentiate componentwise: \(\mathbf{r}'(t)=\left\langle e^{t}, (t+1)e^{t}\right\rangle\). 2. At \(t=0\), the derivative is \(\left\langle 1, 1\right\rangle\), with magnitude \(\sqrt{2}\). 3. At \(t=1\), the derivative is \(\left\langle e, 2e\right\rangle\), with magnitude \(e\sqrt{5}\). 4. Both vectors point right and upward, but the vector at \(t=1\) is longer and has a larger vertical-to-horizontal component ratio.

At \(t=0\): \(\left\langle 1, 1\right\rangle\) At \(t=1\): \(\left\langle e, 2e\right\rangle\) Both point right and upward; the second vector is longer and steeper.

- Expand or use the product rule on the first component before simplifying its derivative. - Differentiate the cubic second component separately. - Substitute \(t=-1\) carefully into the odd and even powers.

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle 4t^{3}-2, 3t^{2}\right\rangle\). 2. At \(t=-1\), the derivative is \(\mathbf{r}'(-1)=\left\langle -6, 3\right\rangle\).

\(\mathbf{r}'(t)=\left\langle 4t^{3}-2, 3t^{2}\right\rangle\); \(\mathbf{r}'(-1)=\left\langle -6, 3\right\rangle\)

- Apply the chain rule to both multiple-angle trigonometric components. - Keep each inner-rate factor with its corresponding component. - Evaluate the exact sine and cosine values after substituting \(t=\frac{\pi}{2}\).

1. Differentiate each component: \(\mathbf{r}'(t)=\left\langle -3\sin(3t), 2\cos(2t)\right\rangle\). 2. At \(t=\frac{\pi}{2}\), the derivative is \(\mathbf{r}'\left(\frac{\pi}{2}\right)=\left\langle 3, -2\right\rangle\).

\(\mathbf{r}'(t)=\left\langle -3\sin(3t), 2\cos(2t)\right\rangle\); \(\mathbf{r}'\left(\frac{\pi}{2}\right)=\left\langle 3, -2\right\rangle\)

- Use the logarithm derivative in the first component. - Apply the product rule to \(te^{t}\) in the second component. - Evaluate the ordered derivative vector at \(t=1\).

1. On the domain \(t>0\), differentiate each component: \(\mathbf{r}'(t)=\left\langle \frac{1}{t},(t+1)e^{t}\right\rangle\). 2. At \(t=1\), the derivative is \(\mathbf{r}'(1)=\left\langle 1,2e\right\rangle\).

\(\mathbf{r}'(t)=\left\langle \frac{1}{t},(t+1)e^{t}\right\rangle\) for \(t>0\); \(\mathbf{r}'(1)=\left\langle 1,2e\right\rangle\)

- Evaluate \(\mathbf{r}(1)\) to obtain the point on the tangent line. - Evaluate \(\mathbf{r}'(1)\) to obtain a direction vector. - Use a new scalar parameter \(s\) in point-plus-direction vector form.

1. The point on the curve is \(\mathbf{r}(1)=\left\langle 1, 1\right\rangle\). 2. A tangent direction is \(\mathbf{r}'(1)=\left\langle 2, 3\right\rangle\). 3. Therefore, the tangent line is \(\ell(s)=\left\langle 1, 1\right\rangle+s\left\langle 2, 3\right\rangle\).

\(\ell(s)=\left\langle 1, 1\right\rangle+s\left\langle 2, 3\right\rangle\)

- Find the point on the unit circle at the specified parameter value. - Differentiate the vector function and evaluate the tangent direction there. - Write the line as the point plus \(s\) times the direction vector.

1. The point on the curve is \(\mathbf{r}\left(\frac{\pi}{2}\right)=\left\langle 0, 1\right\rangle\). 2. A tangent direction is \(\mathbf{r}'\left(\frac{\pi}{2}\right)=\left\langle -1, 0\right\rangle\). 3. Therefore, the tangent line is \(\ell(s)=\left\langle 0, 1\right\rangle+s\left\langle -1, 0\right\rangle\).

\(\ell(s)=\left\langle 0, 1\right\rangle+s\left\langle -1, 0\right\rangle\)

- Evaluate both vector components at \(t=0\) to find the tangent point. - Use the derivative vector at \(t=0\) as the line direction. - Keep \(s\) distinct from the curve parameter \(t\).

1. The point on the curve is \(\mathbf{r}(0)=\left\langle 1, 0\right\rangle\). 2. A tangent direction is \(\mathbf{r}'(0)=\left\langle 1, 1\right\rangle\). 3. Therefore, the tangent line is \(\ell(s)=\left\langle 1, 0\right\rangle+s\left\langle 1, 1\right\rangle\).

\(\ell(s)=\left\langle 1, 0\right\rangle+s\left\langle 1, 1\right\rangle\)

- Evaluate the reciprocal coordinate expressions at \(t=1\). - Differentiate both components and evaluate the resulting direction vector. - Use point-plus-direction form with a separate line parameter.

1. The point on the curve is \(\mathbf{r}(1)=\left\langle 2, 0\right\rangle\). 2. A tangent direction is \(\mathbf{r}'(1)=\left\langle 0, 2\right\rangle\). 3. Therefore, the tangent line is \(\ell(s)=\left\langle 2, 0\right\rangle+s\left\langle 0, 2\right\rangle\).

\(\ell(s)=\left\langle 2, 0\right\rangle+s\left\langle 0, 2\right\rangle\)

- Compute \(\mathbf{r}(2)\) for the point on the curve. - Compute \(\mathbf{r}'(2)\) for the tangent direction. - Introduce \(s\) as the parameter of the tangent line, not the original curve.

1. The point on the curve is \(\mathbf{r}(2)=\left\langle 2, 4\right\rangle\). 2. A tangent direction is \(\mathbf{r}'(2)=\left\langle 9, 4\right\rangle\). 3. Therefore, the tangent line is \(\ell(s)=\left\langle 2, 4\right\rangle+s\left\langle 9, 4\right\rangle\).

\(\ell(s)=\left\langle 2, 4\right\rangle+s\left\langle 9, 4\right\rangle\)

- Interpret the sign of each derivative component separately. - Solve the horizontal and vertical sign conditions as inequalities. - Intersect the two solution sets.

1. The x-coordinate increases when \(t^2-4>0\), which means \(t<-2\) or \(t>2\). 2. The y-coordinate decreases when \(2t+1<0\), which means \(t<-\frac{1}{2}\). 3. Both conditions hold only when \(t<-2\).

\(t<-2\)

- Treat each component as an accumulated scalar function. - Differentiate using the variable upper limit. - Evaluate the resulting vector only after differentiating.

1. Differentiate each accumulated component using the Fundamental Theorem of Calculus. 2. This gives \(\mathbf{r}'(t)=\langle e^{t^2},1+t^3\rangle\). 3. Therefore, \(\mathbf{r}'(1)=\langle e,2\rangle\).

\(\langle e,2\rangle\)

- Write a vector equation for the tangent line at the specified parameter. - A point lies on the line only if one line parameter works in both components. - Test the two candidate points separately.

1. The point on the curve is \(\mathbf{r}(1)=(2,1)\), and the tangent vector is \(\mathbf{r}'(1)=\langle1,2\rangle\). 2. The tangent line is \(\langle x,y\rangle=\langle2,1\rangle+s\langle1,2\rangle\). 3. For \(P=(4,5)\), the x-component gives \(s=2\), and the y-component also gives \(s=2\), so \(P\) lies on the line. 4. For \(Q=(5,6)\), the x-component gives \(s=3\), while the y-component gives \(s=\frac{5}{2}\), so \(Q\) does not lie on the line.

Only \(P=(4,5)\) lies on the tangent line.

- A vector-valued derivative should retain one component for each coordinate. - Differentiate the coordinate functions separately. - Compare the type of the student’s scalar result with the requested vector result.

1. Differentiate componentwise: \(\mathbf{r}'(t)=\langle\cos t,-2\sin(2t)\rangle\). 2. At \(t=\frac{\pi}{4}\), the derivative vector is \(\left\langle\frac{\sqrt{2}}{2},-2\right\rangle\). 3. The student reported the magnitude of this vector, since \(\sqrt{\frac{1}{2}+4}=\frac{3\sqrt{2}}{2}\), rather than the vector itself.

The student found the magnitude instead of the derivative vector. The correct derivative is \(\mathbf{r}'\left(\frac{\pi}{4}\right)=\left\langle\frac{\sqrt{2}}{2},-2\right\rangle\).

- Differentiate the transformed coordinates in their displayed order. - Read the component rates from the supplied derivative vector. - Keep track of the negative sign in the first transformed coordinate.

1. Differentiate componentwise: \(\mathbf{p}'(t)=\langle-y'(t),x'(t)\rangle\). 2. From \(\mathbf{r}'(3)=\langle4,-2\rangle\), \(x'(3)=4\) and \(y'(3)=-2\). 3. Therefore, \(\mathbf{p}'(3)=\langle2,4\rangle\).

\(\langle2,4\rangle\)

- Follow the labeled parameter positions in the order \(A\), \(B\), \(C\). - Track horizontal position separately from vertical position. - Look for a point where one component changes from decreasing to increasing.

1. The labeled positions progress from \(A=(-1,1)\) through \(B=(0,0)\) to \(C=(1,1)\). 2. The x-coordinate increases throughout the motion. 3. The y-coordinate decreases from \(t=0\) to \(t=1\), reaches its minimum at \(B\), and then increases from \(t=1\) to \(t=2\).

The x-component increases throughout. The y-component decreases until \(t=1\), then increases.

- Compare the first components to see whether one parameter can be written in terms of the other. - Substitute that parameter relationship into the remaining component. - Check whether the parameter relationship preserves or reverses order.

1. Set \(t=2s\). As \(s\) runs from \(0\) to \(1\), \(t\) runs from \(0\) to \(2\). 2. Substituting \(t=2s\) into \(\mathbf r\) gives \(\mathbf r(2s)=\langle2s,(2s)^2\rangle=\langle2s,4s^2\rangle=\mathbf q(s)\). 3. Because \(t=2s\) increases with \(s\), both parameterizations trace the same segment in the same direction.

Yes. They trace the same curve segment in the same direction.

- Differentiate the curve to obtain its tangent vector as a function of \(t\). - Use proportional components or a zero determinant to impose parallelism. - Reject any parameter value that makes the tangent vector the zero vector.

1. The tangent vector is \(\mathbf{r}'(t)=\left\langle 2t, 3t^{2}\right\rangle\). 2. Parallel vectors have zero determinant, so \((2t)(3)-(3t^{2})(2)=6t(1-t)=0\). The candidate \(t=0\) gives the zero vector, so it is not a tangent direction. 3. At \(t=1\), the point is \(\left\langle 1, 1\right\rangle\), and the tangent vector is \(\left\langle 2, 3\right\rangle\).

\(t=1\); point \(\left\langle 1, 1\right\rangle\)

- Parallel vectors need not be equal; compare their directions using both components. - A determinant or component-ratio condition can encode the direction requirement without introducing an unnecessary scale variable. - Check every parameter solution by confirming the resulting tangent vector is nonzero and parallel to the target.

1. The tangent vector is \(\mathbf r'(t)=\langle2t+1,\ 3t^2-1\rangle\). 2. Parallelism requires proportional components: \(2(2t+1)-3(3t^2-1)=0\). 3. This gives \(9t^2-4t-5=0\), so \(t=1\) or \(t=-\frac59\). 4. The corresponding points are \((2,0)\) and \((-\frac{20}{81},\frac{280}{729})\), respectively.

\(t=1\) at \((2,0)\), and \(t=-\frac59\) at \((-\frac{20}{81},\frac{280}{729})\).

- Differentiate the circular vector function to obtain the tangent direction. - Solve for both same-direction and opposite-direction multiples of \(\left\langle1,1\right\rangle\). - Check all solutions in the stated interval and evaluate the corresponding points.

1. The tangent vector is \(\mathbf{r}'(t)=\left\langle -\sin(t), \cos(t)\right\rangle\). 2. Parallelism requires \(-\sin(t)=\cos(t)\), so the parameter values are \(t=\frac{3\pi}{4}\) and \(t=\frac{7\pi}{4}\). 3. The corresponding points are \(\left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle\) and \(\left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right\rangle\).

\(t=\frac{3\pi}{4}\), point \(\left\langle -\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right\rangle\); \(t=\frac{7\pi}{4}\), point \(\left\langle \frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\right\rangle\)

- A perpendicularity condition should involve both components of each vector. - Differentiate the position vector before applying the direction condition. - After solving for the parameter, substitute it into the original vector function, not the tangent vector, to find the point.

1. The tangent vector is \(\mathbf r'(t)=\langle2t-2,\ 2t+2\rangle\). 2. Perpendicularity requires \(\langle2t-2,2t+2\rangle\cdot\langle2,-1\rangle=0\). 3. This gives \(2(2t-2)-(2t+2)=2t-6=0\), so \(t=3\). 4. The point is \(\mathbf r(3)=\langle3,15\rangle\).

\(t=3\); point \((3,15)\)

- Differentiate both exponential components, including the sign from \(e^{-t}\). - Use a zero dot product with \(\left\langle1,1\right\rangle\) to impose perpendicularity. - Evaluate the original vector function at the solution.

1. The tangent vector is \(\mathbf{r}'(t)=\left\langle e^{t}, -e^{-t}\right\rangle\). 2. Perpendicular vectors have dot product zero, so \(e^{t}-e^{-t}=0\). Thus, \(e^{2t}=1\), giving \(t=0\). 3. At this value, the point is \(\left\langle 1, 1\right\rangle\), and the tangent vector is \(\left\langle 1, -1\right\rangle\).

\(t=0\); point \(\left\langle 1, 1\right\rangle\)

- Vector equality means every corresponding component must agree. - Differentiate before evaluating at the stated parameter. - Treat the two component equations as simultaneous requirements, not as a single scalar check.

1. Differentiate componentwise: \(\mathbf r'(t)=\langle2at,\ 3bt^2\rangle\). 2. At \(t=1\), vector equality gives \(\langle2a,3b\rangle=\langle4,3\rangle\). 3. Matching both components gives \(a=2\) and \(b=1\).

\(a=2\), \(b=1\)

- Evaluate each parameterization before comparing their tangent information. - Scalar multiples determine the same tangent line, but the sign affects orientation. - Use either nonzero tangent vector to obtain the line’s slope.

1. Both parameter values produce the point \((1,1)\). 2. The tangent vectors are \(\mathbf{r}_1'(1)=\langle1,2\rangle\) and \(\mathbf{r}_2'(1)=\langle-1,-2\rangle\). 3. The vectors are negative scalar multiples, so the same curve is traced in opposite directions at the point. 4. Both vectors determine slope \(2\), so the common tangent line is \(y-1=2(x-1)\).

The tangent vectors are \(\langle1,2\rangle\) and \(\langle-1,-2\rangle\), so the directions are opposite. The common tangent line is \(y-1=2(x-1)\).

- Differentiate the two vector-valued functions independently. - Equality of vectors requires equality of corresponding components. - Check the final pair in both component equations.

1. The derivative vectors are \(\mathbf{r}'(t)=\langle2t,3t^2\rangle\) and \(\mathbf{q}'(u)=\langle2,6u\rangle\). 2. Equality of first components gives \(2t=2\), so \(t=1\). 3. Equality of second components then gives \(3=6u\), so \(u=\frac{1}{2}\).

\((t,u)=\left(1,\frac{1}{2}\right)\)

- Treat the polynomial factor as a scalar function multiplying a vector function. - Both the scalar and vector factors contribute to the derivative. - Substitute the supplied vector data only after differentiating.

1. Differentiate the scalar–vector product: \(\mathbf{p}'(t)=2t\mathbf{r}(t)+(t^2+1)\mathbf{r}'(t)\). 2. At \(t=1\), \(\mathbf{p}'(1)=2\langle2,-1\rangle+2\langle3,4\rangle\). 3. Therefore, \(\mathbf{p}'(1)=\langle10,6\rangle\).

\(\langle10,6\rangle\)

- A vertical vector has no horizontal component. - The sign of the remaining component determines whether it points up or down. - Evaluate the position only after selecting the correct parameter value.

1. The derivative vector is \(\mathbf{r}'(t)=\langle3t^2-3,2t+1\rangle\). 2. A vertical tangent vector has first component \(0\), so \(t=-1\) or \(t=1\). 3. At \(t=-1\), the second component is \(-1\), so the vector points downward. At \(t=1\), it is \(3\), so the vector points upward. 4. The requested point is \(\mathbf{r}(1)=(-2,2)\).

\((-2,2)\)

- Treat the vector function as the outer function in a composition. - Match the supplied derivative vector to the inner function’s value. - Scale the outer derivative vector by the inner rate.

1. The chain rule gives \(\frac{d}{dt}[\mathbf{r}(g(t))]=\mathbf{r}'(g(t))g'(t)\). 2. At \(t=1\), this is \(\mathbf{r}'(2)(4)\). 3. Therefore, the derivative vector is \(4\langle3,-1\rangle=\langle12,-4\rangle\).

\(\langle12,-4\rangle\)

- Differentiate the two components independently. - The components require different scalar derivative rules. - Substitute the tabulated values only after writing each derivative.

1. The first component derivative is \(f'g+fg'=3(-1)+2(4)=5\). 2. The second component derivative is \(\frac{f'g-fg'}{g^2}=\frac{3(-1)-2(4)}{(-1)^2}=-11\). 3. Therefore, \(\mathbf{r}'(1)=\langle5,-11\rangle\).

\(\langle5,-11\rangle\)

- Differentiate the vector function before using either data point. - Equate corresponding components at each parameter value. - Use the simpler \(t=0\) condition first.

1. The derivative is \(\mathbf{r}'(t)=\langle2at+b,3ct^2\rangle\). 2. From \(\mathbf{r}'(0)=\langle2,0\rangle\), \(b=2\). 3. From the first component at \(t=1\), \(2a+2=8\), so \(a=3\). 4. From the second component at \(t=1\), \(3c=6\), so \(c=2\).

\(a=3\), \(b=2\), and \(c=2\)

- Find both the point and derivative vector at the specified parameter. - Use a new parameter for the tangent line. - Impose the y-axis condition on the line’s x-component.

1. The point on the curve is \(\mathbf{r}(1)=\langle2,0\rangle\). 2. The derivative vector is \(\mathbf{r}'(1)=\langle2,2\rangle\). 3. A vector equation of the tangent line is \(\langle x,y\rangle=\langle2,0\rangle+s\langle2,2\rangle\). 4. On the y-axis, \(x=0\), so \(2+2s=0\) and \(s=-1\). The intersection is \((0,-2)\).

\((0,-2)\)

- Use the magnitude condition on the derivative vector before applying the direction condition. - Squaring the magnitude avoids an unnecessary radical. - The sign of the vertical component selects the correct parameter value.

1. The derivative vector is \(\mathbf{r}'(t)=\langle1,2t\rangle\). 2. Its squared magnitude is \(1+4t^2\). Setting this equal to \(5\) gives \(t=\pm1\). 3. The vector points upward when its second component is positive, so \(t=1\). 4. The corresponding point is \((1,1)\).

\((1,1)\)

- Compare the position and derivative vectors using a perpendicularity condition. - Simplify the resulting polynomial completely. - Check which factors can be zero for real parameter values.

1. The derivative vector is \(\mathbf{r}'(t)=\langle2t,1\rangle\). 2. Perpendicularity requires \(\mathbf{r}(t)\cdot\mathbf{r}'(t)=0\). 3. The dot product is \(2t(t^2+1)+t=t(2t^2+3)\). 4. The only real solution is \(t=0\).

\(t=0\)

- A zero derivative vector requires both components to vanish simultaneously. - Find the zero set of each component separately. - Look for common parameter values rather than combining unrelated roots.

1. The derivative vector is \(\mathbf{r}'(t)=\langle3t(t-2),2(t-1)\rangle\). 2. The first component is zero at \(t=0\) or \(t=2\). 3. The second component is zero only at \(t=1\). 4. No parameter value makes both components zero, so the curve has no stationary parameter values.

No stationary parameter values exist.

- Differentiate the scalar–vector product in the final term. - Keep the three vector contributions separate until substitution. - Combine corresponding components at the end.

1. Differentiate to obtain \(\mathbf{p}'(t)=2\mathbf{r}'(t)-\mathbf{q}(t)-t\mathbf{q}'(t)\). 2. At \(t=2\), \(\mathbf{p}'(2)=2\langle1,-3\rangle-\langle4,1\rangle-2\langle-2,5\rangle\). 3. Combining components gives \(\mathbf{p}'(2)=\langle2,-17\rangle\).

\(\langle2,-17\rangle\)

- Differentiate each composition separately before combining the vectors. - Each outer derivative vector is scaled by the corresponding inner rate. - Match the supplied vector data to the inner function values.

1. Apply the chain rule to each composed vector function: \(\mathbf{p}'(t)=\mathbf{r}'(g(t))g'(t)+\mathbf{q}'(h(t))h'(t)\). 2. At \(t=2\), this becomes \(3\langle2,5\rangle-2\langle-1,3\rangle\). 3. Combining components gives \(\mathbf{p}'(2)=\langle8,9\rangle\).

\(\langle8,9\rangle\)

- Find the derivative vector of each curve at its specified parameter. - Use a condition that characterizes perpendicular vectors. - Solve the resulting scalar equation for the unknown parameter.

1. The first tangent vector is \(\mathbf{r}'(1)=\langle2,3\rangle\). 2. The second tangent vector is \(\mathbf{q}'(a)=\langle1,2a-1\rangle\). 3. Perpendicularity requires \(\langle2,3\rangle\cdot\langle1,2a-1\rangle=0\). 4. Thus \(2+3(2a-1)=0\), so \(a=\frac{1}{6}\).

\(a=\frac{1}{6}\)

- A horizontal tangent vector has a zero vertical component. - Check the sign of the horizontal component to determine its direction. - Evaluate the position only after confirming both conditions.

1. The derivative vector is \(\mathbf{r}'(t)=\langle-3t^2+3,2t-4\rangle\). 2. A horizontal tangent vector requires \(2t-4=0\), so \(t=2\). 3. At \(t=2\), the horizontal component is \(-3(2)^2+3=-9<0\), so the vector points left. 4. The corresponding point is \(\mathbf{r}(2)=(-2,-4)\).

\((-2,-4)\)

- Differentiate the scalar-vector product before substituting any data. - At the specified parameter, two scalar quantities are unknown and the vector equation supplies two component equations. - Check the recovered pair in both components of \(\mathbf p'(2)\).

1. The product rule gives \(\mathbf p'(t)=f'(t)\mathbf r(t)+f(t)\mathbf r'(t)\). 2. Let \(A=f'(2)\) and \(B=f(2)\). Then \(\langle14,-1\rangle=A\langle1,-2\rangle+B\langle4,1\rangle\). 3. Matching components gives \(A+4B=14\) and \(-2A+B=-1\). 4. Solving the system gives \(A=2\) and \(B=3\).

\(f(2)=3\) and \(f'(2)=2\)

- Compare the position and tangent vectors at the same parameter value. - Use a condition for two planar vectors to be parallel. - Simplify the resulting scalar equation before solving.

1. The tangent vector is \(\mathbf{r}'(t)=\langle1,2t\rangle\). 2. Parallel vectors in the plane have determinant \(0\), so \(t(2t)-(t^2+1)(1)=0\). 3. This simplifies to \(t^2-1=0\). 4. Therefore, \(t=-1\) or \(t=1\).

\(t=-1\) or \(t=1\)

- Differentiate both coordinate functions. - Translate “equal components” into an equation. - Use the sign condition to select one of the solutions.

1. The derivative vector is \(\mathbf{r}'(t)=\langle2t,3t^2-3\rangle\). 2. Equal components require \(2t=3t^2-3\), or \(3t^2-2t-3=0\). 3. The solutions are \(t=\frac{1\pm\sqrt{10}}{3}\). 4. Only \(t=\frac{1-\sqrt{10}}{3}\) makes the common component \(2t\) negative.

\(t=\frac{1-\sqrt{10}}{3}\)

- At an endpoint, examine the appropriate one-sided difference quotient. - Consider the limits of the two components separately. - A derivative vector must have finite component limits.

1. The one-sided difference quotient is \(\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h}=\left\langle\frac{1}{\sqrt{h}},\sqrt{h}\right\rangle\) for \(h>0\). 2. As \(h\to0^+\), the first component increases without bound while the second component approaches \(0\). 3. Because the difference quotient does not approach a finite vector, \(\mathbf{r}'(0)\) does not exist as a finite derivative vector.

\(\mathbf{r}'(0)\) does not exist as a finite one-sided derivative vector.

- Use the sign of a dot product to classify the angle between two vectors. - Reduce the condition to a quadratic inequality. - Use the roots and the leading coefficient to determine the solution intervals.

1. The tangent vector is \(\mathbf{r}'(t)=\langle2t,3t^2-3\rangle\). 2. The angle is acute when \(\mathbf{r}'(t)\cdot\langle1,1\rangle>0\). 3. This gives \(3t^2+2t-3>0\). Its zeros are \(t=\frac{-1\pm\sqrt{10}}{3}\). 4. Because the quadratic opens upward, the inequality holds for \(t<\frac{-1-\sqrt{10}}{3}\) or \(t>\frac{-1+\sqrt{10}}{3}\).

\(t<\frac{-1-\sqrt{10}}{3}\) or \(t>\frac{-1+\sqrt{10}}{3}\)

- The argument \(-t\) contributes an additional scalar derivative. - Keep track of the sign introduced by the inner function. - Combine the two derivative vectors componentwise.

1. Differentiate the second term with the chain rule: \(\mathbf{p}'(t)=\mathbf{r}'(t)-\mathbf{r}'(-t)\). 2. At \(t=2\), \(\mathbf{p}'(2)=\langle3,-1\rangle-\langle-2,4\rangle\). 3. Therefore, \(\mathbf{p}'(2)=\langle5,-5\rangle\).

\(\langle5,-5\rangle\)

- Evaluate the derivative vector of each parameterization at the stated values. - Compare whether one vector is a positive or negative scalar multiple of the other. - The sign of the scalar multiple determines orientation.

1. The derivative vectors are \(\mathbf{r}'(t)=\langle2t,3t^2\rangle\) and \(\mathbf{q}'(s)=\langle4s^3,6s^5\rangle\). 2. At \(t=1\), \(\mathbf{r}'(1)=\langle2,3\rangle\). 3. At \(s=1\), \(\mathbf{q}'(1)=\langle4,6\rangle=2\langle2,3\rangle\), so the curves travel in the same direction. 4. At \(s=-1\), \(\mathbf{q}'(-1)=\langle-4,-6\rangle=-2\langle2,3\rangle\), so the curves travel in opposite directions.

At \(t=1\) and \(s=1\), the directions are the same. At \(t=1\) and \(s=-1\), the directions are opposite.

- Continuity and differentiability impose separate vector conditions. - Match the position vectors first, then match the one-sided derivative vectors. - Solve the simpler derivative condition before returning to continuity.

1. Continuity requires the one-sided position vectors to agree: \(\langle1,a+b\rangle=\langle1,1\rangle\), so \(a+b=1\). 2. The left derivative at \(t=1\) is \(\langle2,a\rangle\), and the right derivative is \(\langle2,2\rangle\). 3. Differentiability requires \(a=2\). 4. Substituting into \(a+b=1\) gives \(b=-1\).

\(a=2\) and \(b=-1\)

- Each accumulated component has a nontrivial upper limit. - Differentiate the outer accumulation and then multiply by the upper-limit rate. - Simplify the logarithm before evaluating.

1. Apply the Fundamental Theorem of Calculus and the chain rule to each component. 2. The first component derivative is \(e^{t^4}(2t)\). 3. The second component derivative is \(\ln(e^t)e^t=te^t\). 4. At \(t=1\), \(\mathbf{r}'(1)=\langle2e,e\rangle\).

\(\langle2e,e\rangle\)

- Rewrite the squared magnitude as a dot product of the vector with itself. - Differentiate before substituting the parameter value. - The resulting condition is a scalar equation in the unknown constant.

1. For a vector function, \(\frac{d}{dt}\|\mathbf{r}(t)\|^2=2\mathbf{r}(t)\cdot\mathbf{r}'(t)\). 2. At \(t=1\), \(\mathbf{r}_a(1)=\langle1+a,0\rangle\) and \(\mathbf{r}_a'(1)=\langle2,2\rangle\). 3. Thus the derivative is \(2\langle1+a,0\rangle\cdot\langle2,2\rangle=4(1+a)\). 4. Solving \(4(1+a)=16\) gives \(a=3\).

\(a=3\)

- First compute the vector change over the full parameter interval. - Compare that average vector with the instantaneous derivative vector. - Use a parallel-vector condition and then apply the open-interval restriction.

1. The average velocity vector is \(\frac{\mathbf{r}(2)-\mathbf{r}(0)}{2-0}=\langle2,4\rangle\). 2. The tangent vector is \(\mathbf{r}'(t)=\langle2t,3t^2\rangle\). 3. Parallel vectors satisfy \((2t)(4)-(3t^2)(2)=0\), so \(2t(4-3t)=0\). 4. The solution in \((0,2)\) is \(t=\frac{4}{3}\).

\(t=\frac{4}{3}\)

- Rewrite the absolute-value components separately on the two sides of zero. - Differentiate each side before taking the one-sided limits. - Vector differentiability requires every component derivative to agree.

1. For \(t<0\), \(\mathbf{r}(t)=\langle-t,-t^2\rangle\), so the left-hand derivative at \(0\) is \(\langle-1,0\rangle\). 2. For \(t>0\), \(\mathbf{r}(t)=\langle t,t^2\rangle\), so the right-hand derivative at \(0\) is \(\langle1,0\rangle\). 3. The one-sided derivative vectors are unequal, so \(\mathbf{r}'(0)\) does not exist.

The function is not differentiable at \(t=0\). The one-sided derivative vectors are \(\langle-1,0\rangle\) and \(\langle1,0\rangle\).

- Translate the direction and speed statements into two different properties of the same tangent vector. - The sign condition resolves an ambiguity left by the magnitude equation. - Check the final vector against both the perpendicularity and speed requirements.

1. The tangent vector is \(\mathbf r'(t)=\langle a+e^t,\ b+2t\rangle\), so \(\mathbf r'(0)=\langle a+1,b\rangle\). 2. Perpendicularity gives \(a+1+b=0\), so \(b=-(a+1)\). 3. The speed condition gives \((a+1)^2+b^2=8\). Substituting the perpendicularity relation yields \(2(a+1)^2=8\), so \(|a+1|=2\). 4. Since \(\frac{dx}{dt}=a+1>0\), \(a+1=2\). Therefore, \(a=1\) and \(b=-2\).

\(a=1\), \(b=-2\)

- The position condition and tangent condition refer to two different vectors at the same parameter. - Convert the dot product into one scalar equation and the parallelism into another. - Both equations must hold for the same pair \((a,b)\).

1. At \(t=1\), \(\mathbf r(1)=\langle a+1,b+1\rangle\). The dot-product condition gives \(a+b+2=2\), so \(a+b=0\). 2. The tangent vector is \(\mathbf r'(t)=\langle2at+3t^2,b+2t\rangle\), so \(\mathbf r'(1)=\langle2a+3,b+2\rangle\). 3. Parallelism with \(\langle1,1\rangle\) requires \(2a+3=b+2\), or \(b=2a+1\). 4. Solving with \(a+b=0\) gives \(a=-\frac13\) and \(b=\frac13\).

\(a=-\frac13\), \(b=\frac13\)

- At the special angle, slope depends on a ratio of the two tangent-vector components, while speed depends on their magnitude. - Use the positivity conditions when taking square roots. - Check both the direction ratio and the speed after solving.

1. The tangent vector is \(\mathbf r'(t)=\langle-a\sin(t),\ b\cos(t)\rangle\). 2. At \(t=\frac{\pi}{4}\), the slope is \(-\frac{b}{a}\). Thus \(-\frac{b}{a}=-\frac32\), so \(b=\frac32a\). 3. The speed there is \(\sqrt{\frac{a^2+b^2}{2}}=\sqrt{13}\), so \(a^2+b^2=26\). 4. Substituting \(b=\frac32a\) gives \(\frac{13}{4}a^2=26\), hence \(a=2\sqrt{2}\) because \(a>0\). Therefore, \(b=3\sqrt{2}\).

\(a=2\sqrt{2}\), \(b=3\sqrt{2}\)

- Represent the segment from the origin and the tangent direction as vectors. - Use a vector relationship that characterizes perpendicular directions. - Convert every valid parameter value back to a point on the curve.

1. The position vector is \(\langle t^2-1,t\rangle\), and a tangent vector is \(\langle2t,1\rangle\). 2. Perpendicular vectors have zero dot product, so \(2t(t^2-1)+t=0\). 3. This simplifies to \(t(2t^2-1)=0\), giving \(t=0\) and \(t=\pm\frac{1}{\sqrt{2}}\). 4. The corresponding points are \((-1,0)\), \(\left(-\frac{1}{2},\frac{1}{\sqrt{2}}\right)\), and \(\left(-\frac{1}{2},-\frac{1}{\sqrt{2}}\right)\).

\((-1,0)\), \(\left(-\frac{1}{2},\frac{1}{\sqrt{2}}\right)\), and \(\left(-\frac{1}{2},-\frac{1}{\sqrt{2}}\right)\)

- Express the point and tangent vector at the specified parameter in terms of the unknown constant. - Use a second parameter for the tangent line. - Apply the target point’s two coordinate conditions separately.

1. At \(t=1\), the curve passes through \((1+a,1)\), and its tangent vector is \(\langle2+a,3\rangle\). 2. A point on the tangent line has the form \(\langle1+a,1\rangle+s\langle2+a,3\rangle\). 3. Reaching y-coordinate \(4\) requires \(1+3s=4\), so \(s=1\). 4. The x-coordinate condition is \((1+a)+(2+a)=0\), which gives \(a=-\frac{3}{2}\).

\(a=-\frac{3}{2}\)

- Differentiate both coordinate relations with respect to the parameter. - Remember that both coordinates depend on the parameter. - Substitute the known point before solving the resulting linear system.

1. Differentiate \(x^2+ty=3\) to get \(2xx'+y+ty'=0\). 2. Differentiate \(x+ty^2=5\) to get \(x'+y^2+2tyy'=0\). 3. At \(t=1\), \(x=1\), and \(y=2\), the equations become \(2x'+y'=-2\) and \(x'+4y'=-4\). 4. Solving gives \(x'=-\frac{4}{7}\) and \(y'=-\frac{6}{7}\), so \(\mathbf{r}'(1)=\left\langle-\frac{4}{7},-\frac{6}{7}\right\rangle\).

\(\left\langle-\frac{4}{7},-\frac{6}{7}\right\rangle\)

- Use the derivative definition at the specially defined parameter value. - Bound each oscillating component rather than trying to assign a limit to the trigonometric factor. - A componentwise limit determines the vector limit.

1. The difference quotient at \(0\) is \(\frac{\mathbf{r}(h)-\mathbf{r}(0)}{h}=\langle h\sin(1/h),h\cos(1/h)\rangle\). 2. Each component has absolute value at most \(|h|\). 3. Both components therefore approach \(0\) as \(h\to0\). 4. The derivative exists and equals \(\mathbf{r}'(0)=\langle0,0\rangle\).

\(\mathbf{r}'(0)=\langle0,0\rangle\)