Free math worksheets

- Which listed rate describes vertical change, and which describes horizontal change? - Think of slope as vertical change per horizontal change. - Check that your result keeps the sign information from both rates.

1. Use the two listed component rates at the same parameter value. 2. Their quotient is \(\frac{dy}{dx}=\frac{5}{-3}=-\frac{5}{3}\).

\(\frac{dy}{dx}=-\frac{5}{3}\)

- Compare the rate of change of \(y\) with the rate of change of \(x\). - What sign should a tangent slope have when \(x\) increases while \(y\) decreases? - Check that your result describes change in \(y\) per unit change in \(x\), not per unit change in \(t\).

1. The slope with respect to \(x\) is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{-6}{4}=-\frac{3}{2}\). 2. Because the slope is negative, the tangent falls as \(x\) increases.

\(\frac{dy}{dx}=-\frac{3}{2}\); the tangent falls as \(x\) increases.

- Read the sign of each component graph's slope at the marked parameter value. - Decide what happens to \(y\) while \(x\) is increasing. - You only need the sign of the ratio, not the exact component slopes.

1. At \(t=0\), the graph of \(x(t)\) is increasing, so \(\frac{dx}{dt}>0\). 2. At \(t=0\), the graph of \(y(t)\) is decreasing, so \(\frac{dy}{dt}<0\). 3. Therefore, \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}<0\).

\(\frac{dy}{dx}\) is negative.

- What two coordinate rates determine the direction of a parametric curve at one instant? - Find how each coordinate changes with \(t\) before using the requested parameter value. - Check that the horizontal component rate permits a finite tangent slope at \(t=1\).

1. Differentiate the components: \(\frac{dx}{dt}=2t + 2\) and \(\frac{dy}{dt}=3t^{2}-3\). 2. At \(t=1\), the component rates are \(\frac{dx}{dt}=4\) and \(\frac{dy}{dt}=0\). 3. Their quotient gives \(\frac{dy}{dx}=0\).

\(\frac{dy}{dx}=0\)

- Think about what the tangent slope should compare when both coordinates depend on the same parameter. - Differentiate each coordinate before evaluating anything at \(t=0\). - After finding the two local rates, check which one corresponds to horizontal change.

1. Differentiate the components: \(\frac{dx}{dt}=e^{t}+1\) and \(\frac{dy}{dt}=e^{t}-2\). 2. At \(t=0\), the component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=-1\). 3. Their quotient gives \(\frac{dy}{dx}=-\frac{1}{2}\).

\(\frac{dy}{dx}=-\frac{1}{2}\)

- At the requested parameter, the tangent direction comes from how \(x\) and \(y\) are changing together. - Pay special attention to the signs when differentiating the trigonometric terms. - Use exact trigonometric values and then decide how the two component rates combine into slope.

1. Differentiate the components: \(\frac{dx}{dt}=\cos(t)+1\) and \(\frac{dy}{dt}=-\sin(t)-1\). 2. At \(t=\frac{\pi}{2}\), the component rates are \(\frac{dx}{dt}=1\) and \(\frac{dy}{dt}=-2\). 3. Their quotient gives \(\frac{dy}{dx}=-2\).

\(\frac{dy}{dx}=-2\)

- Before differentiating, note the domain restriction created by the logarithm. - Work out the two coordinate rates carefully, especially the reciprocal term. - At \(t=1\), ask whether the horizontal rate is nonzero and what that means for a finite tangent slope.

1. Differentiate the components: \(\frac{dx}{dt}=2t + \frac{1}{t}\) and \(\frac{dy}{dt}=3 + \frac{1}{t^{2}}\). 2. At \(t=1\), the component rates are \(\frac{dx}{dt}=3\) and \(\frac{dy}{dt}=4\). 3. Their quotient gives \(\frac{dy}{dx}=\frac{4}{3}\).

\(\frac{dy}{dx}=\frac{4}{3}\)

- The tangent direction is determined by the two instantaneous coordinate changes, not by the coordinates themselves. - Differentiate the product in \(y\) before substituting the negative parameter value. - Once the two rates are known, check whether the horizontal rate allows an ordinary finite slope.

1. Differentiate the components: \(\frac{dx}{dt}=3t^{2}-1\) and \(\frac{dy}{dt}=2t+4\). 2. At \(t=-1\), the component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2\). 3. Their quotient gives \(\frac{dy}{dx}=1\).

\(\frac{dy}{dx}=1\)

- The given slope describes how the vertical and horizontal component rates compare. - Decide which component rate is unknown before writing a relationship among the three quantities. - Check whether the sign of your result is consistent with a negative tangent slope and a positive horizontal rate.

1. Relate the tangent slope to the two component rates. 2. Solve \(-\frac{1}{2}=\frac{dy/dt}{4}\) to obtain \(\frac{dy}{dt}=-2\).

\(\frac{dy}{dt}=-2\)

- Think about how tangent slope compares vertical and horizontal component rates. - Here the vertical rate is known, so determine what horizontal rate would make the stated slope possible. - Substitute your candidate rate back into the slope relationship as a check.

1. Relate the tangent slope to the two component rates. 2. Solve \(4=\frac{6}{dx/dt}\) to obtain \(\frac{dx}{dt}=\frac{3}{2}\).

\(\frac{dx}{dt}=\frac{3}{2}\)

- Compare the vertical and horizontal components of the same tangent direction. - Look for a common factor in the two rates. - Distinguish a vector’s magnitude from the direction it determines.

1. The tangent slope is the ratio of the vertical component rate to the horizontal component rate. 2. Therefore, \(\frac{dy}{dx}=\frac{-2q}{3q}=-\frac{2}{3}\). 3. The common nonzero factor \(q\) cancels, so changing \(q\) changes the magnitude of the component rates but not their direction ratio.

\(\frac{dy}{dx}=-\frac{2}{3}\); the common nonzero factor \(q\) cancels.

- Differentiate each accumulated function with respect to its upper limit. - Evaluate the two component rates at the stated parameter value. - Compare the vertical rate with the horizontal rate.

1. By the Fundamental Theorem of Calculus, \(\frac{dx}{dt}=1+t^2\) and \(\frac{dy}{dt}=3-t\). 2. At \(t=1\), both component rates equal \(2\). 3. Therefore, \(\frac{dy}{dx}=\frac{2}{2}=1\).

\(1\)

- A tangent line needs both a point on the curve and its local direction there. - Use the parameter value separately in the coordinate functions and in the component rates. - Once the point and finite tangent slope are known, use any equivalent line form.

1. The point is \((5, 1)\). 2. Differentiate the components: \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=2\). 3. Using point-slope form, an equation of the tangent line is \(y-1=2(x-5)\).

\(y-1=2(x-5)\)

- First distinguish the direction of the curve from the direction perpendicular to the curve. - Find the point and the tangent direction at the stated parameter before thinking about the normal line. - Check the geometric relationship between the tangent and normal slopes before writing the line.

1. The point is \((0, 9)\). 2. Differentiate the components: \(\frac{dx}{dt}=2t - 2\) and \(\frac{dy}{dt}=3t^{2}\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=6\). 3. The normal slope is the negative reciprocal, \(-\frac{1}{6}\). 4. Using point-slope form, an equation of the normal line is \(y-9=-\frac{1}{6}x\).

\(y-9=-\frac{1}{6}x\)

- Separate the two pieces of information a tangent line needs: its contact point and its direction. - Keep the exact trigonometric values at \(t=\frac{\pi}{2}\) rather than converting them to decimals. - Check the signs of both component rates before deciding whether the tangent rises or falls as \(x\) increases.

1. The point is \(\left(0, 1 + \frac{\pi}{2}\right)\). 2. Differentiate the components: \(\frac{dx}{dt}=-\sin(t)\) and \(\frac{dy}{dt}=1 + \cos(t)\), so the tangent slope at \(t=\frac{\pi}{2}\) is \(\frac{dy}{dx}=-1\). 3. Using point-slope form, an equation of the tangent line is \(y-\left(1 + \frac{\pi}{2}\right)=-x\).

\(y-\left(1 + \frac{\pi}{2}\right)=-x\)

- The reciprocal terms affect the two component derivatives with opposite signs; check those signs carefully. - Determine the tangent direction at the parameter value before converting to the perpendicular direction. - Keep the fractional contact point exact when you write the normal line.

1. The point is \(\left(\frac{5}{2}, \frac{3}{2}\right)\). 2. Differentiate the components: \(\frac{dx}{dt}=1 - \frac{1}{t^{2}}\) and \(\frac{dy}{dt}=1 + \frac{1}{t^{2}}\), so the tangent slope at \(t=2\) is \(\frac{dy}{dx}=\frac{5}{3}\). 3. The normal slope is the negative reciprocal, \(-\frac{3}{5}\). 4. Using point-slope form, an equation of the normal line is \(y-\frac{3}{2}=-\frac{3}{5}\left(x-\frac{5}{2}\right)\).

\(y-\frac{3}{2}=-\frac{3}{5}\left(x-\frac{5}{2}\right)\)

- Find the contact point and the local direction as two separate pieces of the tangent-line problem. - The second coordinate is a product of two functions of \(t\); make sure its rate of change includes both contributions. - After evaluating at \(t=0\), check whether the tangent direction is finite before writing the line.

1. The point is \((1, 0)\). 2. Differentiate the components: \(\frac{dx}{dt}=e^{t}\) and \(\frac{dy}{dt}=e^{t}+te^{t}\), so the tangent slope at \(t=0\) is \(\frac{dy}{dx}=1\). 3. Using point-slope form, an equation of the tangent line is \(y=x-1\).

\(y=x-1\)

- What must be true about the curve's instantaneous vertical change for the tangent to be horizontal? - A candidate is not automatically valid; also consider what the horizontal coordinate is doing there. - After identifying the valid parameter values, return to the original coordinate equations to locate the points.

1. Differentiate: \(\frac{dx}{dt}=2t-4\) and \(\frac{dy}{dt}=3t^{2}-3\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne0\). 3. The valid parameter values are \(t=-1\) and \(t=1\). 4. Substitution gives the points \((5, 2)\) and \((-3, -2)\).

Parameter values: \(t=-1\), \(t=1\) Points: \((5, 2)\), \((-3, -2)\)

- Translate “horizontal tangent” into a statement about the two instantaneous coordinate changes. - Solve the relevant trigonometric condition over the entire interval, not just one quadrant. - Use the other component rate to screen the candidates before finding their coordinates.

1. Differentiate: \(\frac{dx}{dt}=-\sin(t)\) and \(\frac{dy}{dt}=2\cos(2t)\). 2. A horizontal tangent requires \(\frac{dy}{dt}=0\) and \(\frac{dx}{dt}\ne0\). 3. The valid parameter values are \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), and \(t=\frac{7\pi}{4}\). 4. Substitution gives the points \(\left(\frac{\sqrt{2}}{2}, 1\right)\), \(\left(-\frac{\sqrt{2}}{2}, -1\right)\), \(\left(-\frac{\sqrt{2}}{2}, 1\right)\), and \(\left(\frac{\sqrt{2}}{2}, -1\right)\).

Parameter values: \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), \(t=\frac{7\pi}{4}\) Points: \(\left(\frac{\sqrt{2}}{2}, 1\right)\), \(\left(-\frac{\sqrt{2}}{2}, -1\right)\), \(\left(-\frac{\sqrt{2}}{2}, 1\right)\), \(\left(\frac{\sqrt{2}}{2}, -1\right)\)

- For a vertical tangent, ask which coordinate must momentarily stop changing and which must still be changing. - Track all trigonometric solutions over \([0,2\pi]\); periodicity creates several candidates. - Verify the candidates before evaluating the original coordinates.

1. Differentiate: \(\frac{dx}{dt}=2\cos(2t)\) and \(\frac{dy}{dt}=-\sin(t)\). 2. A vertical tangent requires \(\frac{dx}{dt}=0\) and \(\frac{dy}{dt}\ne0\). 3. The valid parameter values are \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), and \(t=\frac{7\pi}{4}\). 4. Substitution gives the points \(\left(1, \frac{\sqrt{2}}{2}\right)\), \(\left(-1, -\frac{\sqrt{2}}{2}\right)\), \(\left(1, -\frac{\sqrt{2}}{2}\right)\), and \(\left(-1, \frac{\sqrt{2}}{2}\right)\).

Parameter values: \(t=\frac{\pi}{4}\), \(t=\frac{3\pi}{4}\), \(t=\frac{5\pi}{4}\), \(t=\frac{7\pi}{4}\) Points: \(\left(1, \frac{\sqrt{2}}{2}\right)\), \(\left(-1, -\frac{\sqrt{2}}{2}\right)\), \(\left(1, -\frac{\sqrt{2}}{2}\right)\), \(\left(-1, \frac{\sqrt{2}}{2}\right)\)

- The given tangent slope imposes a relationship between the two component rates at \(t=1\). - Differentiate while treating \(a\) as a constant, then evaluate the rates at the specified parameter. - After solving for \(a\), check that the horizontal component rate is not zero.

1. The component derivatives are \(\frac{dx}{dt}=a+2t\) and \(\frac{dy}{dt}=3t^{2}-1\). 2. At \(t=1\), set the quotient equal to \(2\): \(\frac{2}{a+2}=2\). 3. Solving gives \(a=-1\).

\(a=-1\)

- A slope of \(1\) means the instantaneous horizontal and vertical changes are related in a particularly simple way. - Find the component rates at \(t=0\) while keeping \(a\) unknown. - Check that the value you obtain produces an ordinary finite tangent slope.

1. The component derivatives are \(\frac{dx}{dt}=a\cos(t)+1\) and \(\frac{dy}{dt}=2-\sin(t)\). 2. At \(t=0\), set the quotient equal to \(1\): \(\frac{2}{a+1}=1\). 3. Solving gives \(a=1\).

\(a=1\)

- Keep the unknown coefficient in the horizontal rate instead of trying to solve from the original coordinates. - Evaluate the two component rates at \(t=0\) and use the stated geometric slope as a constraint between them. - Verify that your value of \(a\) leaves a nonzero horizontal rate.

1. The component derivatives are \(\frac{dx}{dt}=a+e^{t}\) and \(\frac{dy}{dt}=2e^{t}-1\). 2. At \(t=0\), set the quotient equal to \(\frac{1}{2}\): \(\frac{1}{a+1}=\frac{1}{2}\). 3. Solving gives \(a=1\).

\(a=1\)

- The unknown coefficient affects the local horizontal change, while the vertical change is determined once \(t=-1\) is used. - Differentiate first, then interpret the given negative slope as a relationship between the two rates. - Use the sign of the requested slope as a check on your final component rates.

1. The component derivatives are \(\frac{dx}{dt}=a+3t^{2}\) and \(\frac{dy}{dt}=2t-4\). 2. At \(t=-1\), set the quotient equal to \(-1\): \(-\frac{6}{a+3}=-1\). 3. Solving gives \(a=3\).

\(a=3\)

- Compute the two component derivatives independently before judging the student's expression. - Think of ordinary slope as vertical change compared with horizontal change; which rate should play each role here? - After correcting the expression, check where its denominator vanishes.

1. Differentiate the coordinates: \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^{2}-2\). 2. The student reversed the component derivatives. The tangent derivative is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\), not \(\frac{dx/dt}{dy/dt}\). 3. Therefore, \(\frac{dy}{dx}=\frac{3t^{2}-2}{2t}=\frac{3t}{2}-\frac{1}{t}\), for \(t\ne0\).

The student reversed the component derivatives. The correct derivative is \(\frac{dy}{dx}=\frac{3t^{2}-2}{2t}=\frac{3t}{2}-\frac{1}{t}\), for \(t\ne0\).

- Check the reported slope from the two coordinate rates at the stated parameter. - Keep the exact special-angle values rather than estimating them. - Compare the sign as well as the magnitude of your result with the reported value.

1. The component rates are \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=-2\sin(2t)\). 2. At \(t=\frac{\pi}{4}\), these are \(1+\frac{\sqrt2}{2}\) and \(-2\), respectively. 3. Therefore, \(\frac{dy}{dx}=\frac{-2}{1+\frac{\sqrt{2}}{2}}=2\sqrt{2}-4\), which is not \(2\).

The report is incorrect. The tangent slope is \(2\sqrt{2}-4\).

- Find the horizontal and vertical component rates and compare them with the two proposed expressions. - Ordinary slope is vertical change per horizontal change; use that meaning to decide which expression has the correct orientation. - The simplified expression has a parameter restriction; identify where its denominator is zero.

1. Differentiate the coordinates: \(\frac{dx}{dt}=3t^{2}\) and \(\frac{dy}{dt}=2t\). 2. Form the quotient in the correct order: \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2t}{3t^{2}}\). 3. Thus, \(\frac{dy}{dx}=\frac{2}{3t}\) for \(t\ne0\). The expression \(\frac{3t}{2}\) reverses the quotient.

\(\frac{dy}{dx}=\frac{2}{3t}\) for \(t\ne0\), because \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}\). The other expression reverses the quotient.

- Differentiate both exponential coordinates and identify which derivative represents horizontal change and which represents vertical change. - Use the meaning of slope to inspect the student's quotient order before evaluating it. - At \(t=0\), the exponential factors simplify substantially.

1. Differentiate the coordinates: \(\frac{dx}{dt}=2e^{2t}\) and \(\frac{dy}{dt}=-e^{-t}\). 2. The student reversed the quotient. The tangent derivative is \(\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=-\frac{e^{-3t}}{2}\). 3. At \(t=0\), the slope is \(-\frac{1}{2}\).

The student reversed the quotient. The tangent slope at \(t=0\) is \(-\frac{1}{2}\).

- A parameter value must reproduce both coordinates of the given point, not just one of them. - If one coordinate equation gives more than one candidate, use the other coordinate to distinguish them. - After locating the correct parameter, use the local coordinate rates rather than the coordinates themselves to determine tangent slope.

1. From \(t^{2}-1=3\), the candidates are \(t=2\) and \(t=-2\). Only \(t=2\) also gives \(t^{3}-t=6\). 2. The derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^{2}-1\). 3. At \(t=2\), \(\frac{dy}{dx}=\frac{11}{4}\).

\(t=2\); tangent slope: \(\frac{11}{4}\)

- Compare the given coordinates with both parametric coordinate equations. - Check whether one coordinate function is strictly monotonic to justify uniqueness. - Form the tangent slope from the component derivatives at the verified parameter value.

1. Substituting \(t=\pi\) gives \(x=2\pi - 1\) and \(y=\pi\), so \(t=\pi\) produces the point. 2. Because \(\frac{dx}{dt}=2-\sin(t)\ge 1\), the x-coordinate is strictly increasing, so no other parameter value produces the same point. 3. Also, \(\frac{dy}{dt}=1+\cos(t)\). 4. At \(t=\pi\), \(\frac{dy}{dx}=0\).

\(t=\pi\); tangent slope: \(0\)

- Solve one coordinate equation, then test every candidate in the other coordinate. - Differentiate the reciprocal terms with careful signs. - Evaluate the derivative quotient only at the verified parameter value.

1. The x-coordinate equation gives \(t=2\) or \(t=\frac{1}{2}\). Only \(t=2\) also gives \(y=\frac{3}{2}\). 2. The derivatives are \(\frac{dx}{dt}=1 - \frac{1}{t^{2}}\) and \(\frac{dy}{dt}=1 + \frac{1}{t^{2}}\). 3. At \(t=2\), \(\frac{dy}{dx}=\frac{5}{3}\).

\(t=2\); tangent slope: \(\frac{5}{3}\)

- Use a coordinate that identifies the parameter value cleanly, then verify the other coordinate. - The tangent direction depends on how both coordinates change at that same parameter. - Check that the horizontal rate is nonzero before reporting a finite slope.

1. Since \(e^t=1\), the parameter is \(t=0\), and then \(x=0+\sin(0)=0\). 2. The component rates are \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=e^t\). 3. At \(t=0\), the rates are \(2\) and \(1\), so \(\frac{dy}{dx}=\frac12\).

\(t=0\); tangent slope: \(\frac12\)

- Solve the linear x-coordinate equation for \(t\) and substitute into the y-coordinate. - Differentiate the resulting Cartesian equation with respect to \(x\). - Compare that derivative with the parametric derivative at the point corresponding to \(t=2\).

1. From \(x=2t-1\), \(t=\frac{x+1}{2}\). Substituting into \(y=t^{2}+3\) gives \(y=\frac{x^{2}}{4}+\frac{x}{2}+\frac{13}{4}\). 2. Parametrically, \(\frac{dy}{dx}=\frac{2t}{2}=t\), so at \(t=2\) the slope is \(2\). 3. The corresponding point is \((3, 7)\). From the Cartesian equation, \(\frac{dy}{dx}=\frac{x}{2}+\frac{1}{2}\), which also gives \(2\) at \(x=3\).

The Cartesian equation is \(y=\frac{x^{2}}{4}+\frac{x}{2}+\frac{13}{4}\), and both methods give tangent slope \(2\).

- Multiply the two coordinate expressions to eliminate the exponential parameter. - Differentiate the Cartesian relation implicitly. - Evaluate both derivative methods at the point generated by \(t=0\).

1. Multiplying the coordinate equations gives \(xy=e^{t}e^{-t}=1\). 2. Parametrically, \(\frac{dy}{dx}=\frac{-e^{-t}}{e^{t}}=-e^{-2t}\), so at \(t=0\) the slope is \(-1\). 3. The corresponding point is \((1, 1)\). Implicit differentiation of \(xy=1\) gives \(y+x\frac{dy}{dx}=0\), so \(\frac{dy}{dx}=-\frac{y}{x}=-1\) at \((1,1)\).

The Cartesian relation is \(xy=1\), and both methods give tangent slope \(-1\) at \((1,1)\).

- Evaluate the signs of \(\frac{dx}{dt}\) and \(\frac{dy}{dt}\) separately. - Use their quotient to decide whether \(y\) increases or decreases as \(x\) increases. - Use the two individual signs, not the quotient alone, to state the direction of motion.

1. The component rates are \(\frac{dx}{dt}=2t=1\) and \(\frac{dy}{dt}=3t^{2}-3=-\frac{9}{4}\). 2. The tangent slope is \(\frac{dy}{dx}=-\frac{9}{4}\), so the curve is locally decreasing as \(x\) increases. 3. Because \(\frac{dx}{dt}>0\) and \(\frac{dy}{dt}<0\), the motion for increasing \(t\) is to the right and downward.

Tangent slope: \(-\frac{9}{4}\); the curve is locally decreasing as \(x\) increases, and the increasing-\(t\) direction is right and downward.

- Differentiate both coordinates and evaluate the component rates at \(t=\pi\). - Use the quotient of the rates for the tangent slope. - Interpret the signs of the two rates to describe the parameter’s direction of motion.

1. The component rates are \(\frac{dx}{dt}=1-\sin(t)=1\) and \(\frac{dy}{dt}=-1+\cos(t)=-2\) at \(t=\pi\). 2. The tangent slope is \(\frac{dy}{dx}=-2\). 3. Because \(\frac{dx}{dt}>0\) and \(\frac{dy}{dt}<0\), the motion for increasing \(t\) is to the right and downward.

Tangent slope: \(-2\); the increasing-\(t\) direction is right and downward.

- Translate the word “parallel” into a condition on the tangent slope. - Express the tangent slope in terms of the parameter before solving on the stated interval. - Evaluate both coordinate functions after finding the valid parameter value.

1. The component derivatives are \(\frac{dx}{dt}=2t+2\) and \(\frac{dy}{dt}=3t^2-3\). 2. On the given interval, the tangent slope simplifies to \(\frac{dy}{dx}=\frac{3(t-1)}{2}\). 3. Parallel lines have equal slopes, so \(\frac{3(t-1)}{2}=3\), which gives \(t=3\). 4. At \(t=3\), the point is \((15,18)\).

\(t=3\), at the point \((15,18)\)

- First determine the point and tangent direction at the specified parameter value. - Use the tangent equation to locate where it meets each coordinate axis. - Treat the two intercept lengths as the base and height of a right triangle.

1. At \(t=1\), the curve passes through \((2,3)\). 2. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=-3t^2\), so the tangent slope is \(-\frac{3}{2}\). 3. The tangent line is \(y-3=-\frac{3}{2}(x-2)\), or \(y=-\frac{3}{2}x+6\). 4. The intercepts are \((4,0)\) and \((0,6)\), so the triangle’s area is \(\frac{1}{2}(4)(6)=12\).

\(12\) square units

- Express the x-coordinate and tangent slope separately in terms of \(t\). - The slope comes from comparing the two component rates, not from differentiating an eliminated equation. - After finding the parameter, verify that the requested slope really equals the x-coordinate.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t-1\), so \(\frac{dy}{dx}=\frac{2t-1}{2}\). 2. The x-coordinate is \(2t+1\). Setting the tangent slope equal to it gives \(\frac{2t-1}{2}=2t+1\), so \(t=-\frac{3}{2}\). 3. The point is \(\left(-2,\frac{15}{4}\right)\). At this parameter, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), and \(\frac{dy}{dx}=-2\). 4. The tangent line is \(y-\frac{15}{4}=-2(x+2)\), or \(y=-2x-\frac14\).

Point: \(\left(-2,\frac{15}{4}\right)\) Rates: \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), \(\frac{dy}{dx}=-2\) Tangent: \(y=-2x-\frac14\)

- Determine the secant direction from the two endpoint coordinates first. - Compute the component rates before forming the tangent slope. - The desired parameter must be strictly between the two endpoint parameter values.

1. The endpoints are \((2,2)\) and \((6,10)\), so the secant slope is \(\frac{10-2}{6-2}=2\). 2. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), giving tangent slope \(\frac{dy}{dx}=t\). 3. Parallelism requires \(t=2\), which lies between the endpoint parameter values. 4. The point is \((4,5)\), and there \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), and \(\frac{dy}{dx}=2\).

Secant slope: \(2\) Required point: \((4,5)\) at \(t=2\) Rates: \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\)

- Compare both component rates at the specified parameter value. - Think about the direction of a velocity vector whose horizontal component is zero. - A vertical line is written using a constant x-coordinate.

1. At \(t=1\), the point is \((-2,1)\). 2. The component rates are \(\frac{dx}{dt}=3t^2-3=0\) and \(\frac{dy}{dt}=2t=2\). 3. A zero horizontal rate with a nonzero vertical rate gives a vertical tangent, not a horizontal tangent. 4. The tangent-line equation is \(x=-2\).

The student confused a zero horizontal rate with a zero slope. The tangent line is \(x=-2\).

- Find the point and tangent direction at the nearby known parameter value. - Treat the tangent line as a local model of the curve. - Use the small change in x-coordinate to estimate the corresponding change in y-coordinate.

1. At \(t=1\), the point is \((2,2)\). 2. The tangent slope is \(\frac{3t^2+1}{2t}=2\) at \(t=1\). 3. The tangent-line model is \(y-2=2(x-2)\). 4. At \(x=2.04\), the model gives \(y\approx2+2(0.04)=2.08\).

\(y\approx2.08\)

- First write the tangent slope in terms of the parameter. - Use the parameter restriction when solving the x-coordinate equation for \(t\). - Replace the parameter only after simplifying the derivative.

1. The parametric derivative is \(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\). 2. Since \(x=t^2+1\) and \(t>0\), \(t=\sqrt{x-1}\). 3. Substitution gives \(\frac{dy}{dx}=\frac{3}{2}\sqrt{x-1}\).

\(\frac{dy}{dx}=\frac{3}{2}\sqrt{x-1}\), for \(x>1\)

- Determine the tangent line before considering the distance. - Rewrite the line in a standard form with all terms on one side. - Use the coefficients of that line to measure the perpendicular distance from the origin.

1. At \(t=1\), the point is \((2,1)\), and the tangent slope is \(\frac{3}{2}\). 2. The tangent line is \(y-1=\frac{3}{2}(x-2)\), or \(3x-2y-4=0\). 3. The distance from \((0,0)\) to this line is \(\frac{|{-4}|}{\sqrt{3^2+(-2)^2}}=\frac{4}{\sqrt{13}}\).

\(\frac{4}{\sqrt{13}}\)

- First characterize the slope of any line whose two positive axis intercepts are equal. - Compare that slope with the ratio of the curve's component rates. - Verify the resulting line's two intercepts rather than assuming positivity.

1. A line with equal positive axis intercepts has slope \(-1\). 2. Here \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=t\). 3. Setting \(t=-1\) gives the point \((0,2)\), with component rates \(2\) and \(-2\). 4. The tangent line is \(y=-x+2\), whose two positive intercepts are both \(2\).

\(t=-1\) \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-2\), \(\frac{dy}{dx}=-1\) Tangent line: \(y=-x+2\)

- Check that both parameterized pieces approach the same point before comparing tangents. - Form the tangent slope from the component rates on each side of the join. - A single tangent requires the same limiting direction from both sides.

1. Both pieces approach \((0,0)\), so the curve is continuous at the origin. 2. From the left, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2t\), and \(\frac{dy}{dx}=t\to0\). 3. From the right, \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2\), and \(\frac{dy}{dx}=1\). 4. The one-sided tangent slopes differ, so there is no single tangent line at the origin.

Left: \(\frac{dx}{dt}\to2\), \(\frac{dy}{dt}\to0\), \(\frac{dy}{dx}\to0\) Right: \(\frac{dx}{dt}\to2\), \(\frac{dy}{dt}\to2\), \(\frac{dy}{dx}\to1\) Therefore, no single tangent line exists at the origin.

- Differentiate each composed coordinate with respect to the parameter. - Match the given derivative values to the inputs produced at \(t=1\). - Form the tangent slope from the two resulting component rates.

1. The horizontal component rate is \(X'(t)=2t f'(t^2)\), so \(X'(1)=2(1)(4)=8\). 2. The vertical component rate is \(Y'(t)=3g'(3t-1)\), so \(Y'(1)=3(-3)=-9\). 3. Therefore, \(\frac{dY}{dX}=\frac{-9}{8}\).

\(-\frac{9}{8}\)

- First identify the point from the parameter value and the sign condition. - Differentiate each coordinate relation with respect to the shared parameter. - The tangent slope compares the two resulting component rates.

1. At \(t=0\), the positive coordinate values are \(x=2\) and \(y=2\). 2. Differentiating \(x^2+\sin(t)=4\) gives \(2x\frac{dx}{dt}+\cos(t)=0\), so \(\frac{dx}{dt}=-\frac14\) at \(t=0\). 3. Differentiating \(y^2+e^t=5\) gives \(2y\frac{dy}{dt}+e^t=0\), so \(\frac{dy}{dt}=-\frac14\) at \(t=0\). 4. The tangent slope is \(1\), so the tangent line is \(y-2=x-2\).

\(y-2=x-2\)

- Find the parameter value that produces the shared point on each curve before differentiating. - Compute each tangent direction from its own parameter-rate pair. - Sharing a point is not enough for tangency; compare the slopes before writing the common line.

1. The point \((2,1)\) corresponds to \(t=1\) on \(C_1\) and \(u=1\) on \(C_2\). 2. For \(C_1\), \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so at \(t=1\) the slope is \(1\). 3. For \(C_2\), \(\frac{dx}{du}=4u\) and \(\frac{dy}{du}=4\), so at \(u=1\) the slope is also \(1\). 4. The curves share the point and tangent slope, so the common tangent line is \(y-1=x-2\), or \(y=x-1\).

\(C_1\): \(t=1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=2\), slope \(1\) \(C_2\): \(u=1\), \(\frac{dx}{du}=4\), \(\frac{dy}{du}=4\), slope \(1\) Common tangent: \(y=x-1\)

- Begin by finding where the horizontal coordinate has zero instantaneous rate. - If a candidate makes both coordinate rates zero, the usual quotient test is inconclusive; inspect the curve's local behavior there rather than classifying it immediately. - Only after resolving every candidate should you convert the valid parameter value to a point.

1. Differentiate: \(\frac{dx}{dt}=3t^{2}-3\) and \(\frac{dy}{dt}=2t+2\). 2. Solving \(\frac{dx}{dt}=0\) gives the candidates \(t=-1\) and \(t=1\). 3. At \(t=1\), \(\frac{dy}{dt}=4\ne0\), so the curve has a vertical tangent. 4. At \(t=-1\), both first derivatives are zero. Let \(u=t+1\). Then \(x=2-3u^{2}+u^{3}\), \(y=-1+u^{2}\), and \(\frac{dy}{dx}=\frac{2}{-6+3u}\to-\frac{1}{3}\), so the tangent there is not vertical. 5. Substituting \(t=1\) gives the point \((-2, 3)\).

Parameter value: \(t=1\) Point: \((-2, 3)\)

- Start with the condition suggested by the words “horizontal tangent,” but treat its solutions only as candidates. - If both coordinate rates vanish at one candidate, the ordinary slope quotient does not settle the tangent direction; investigate that point separately. - Keep the interval endpoints in your candidate list and verify every case before finding the points.

1. Differentiate: \(\frac{dx}{dt}=1+\cos(t)\) and \(\frac{dy}{dt}=-\sin(t)\). 2. Solving \(\frac{dy}{dt}=0\) gives the candidates \(t=0\), \(t=\pi\), and \(t=2\pi\). 3. At \(t=0\) and \(t=2\pi\), \(\frac{dx}{dt}=2\ne0\), so the tangents are horizontal. 4. At \(t=\pi\), both first derivatives are zero. Let \(u=t-\pi\). Then \(\frac{dy}{dx}=\frac{\sin(u)}{1-\cos(u)}=\cot\left(\frac{u}{2}\right)\), which is unbounded as \(u\to0\), so the tangent there is vertical rather than horizontal. 5. Substitution gives the points \((0, 1)\) and \((2\pi, 1)\).

Parameter values: \(t=0\), \(t=2\pi\) Points: \((0, 1)\), \((2\pi, 1)\)

- Identify every parameter value that produces the stated crossing point. - The curve can have a different tangent direction each time it visits the same point. - Keep the exact radical values when writing the two line equations.

1. Solving \(t^2=3\) gives the two parameter values \(t=-\sqrt{3}\) and \(t=\sqrt{3}\); both make \(y=0\). 2. The component derivatives are \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2-3\). 3. At \(t=-\sqrt{3}\), the tangent slope is \(-\sqrt{3}\). At \(t=\sqrt{3}\), the tangent slope is \(\sqrt{3}\). 4. Using the common point \((3,0)\), the tangent lines are \(y=-\sqrt{3}(x-3)\) and \(y=\sqrt{3}(x-3)\).

\(y=-\sqrt{3}(x-3)\) and \(y=\sqrt{3}(x-3)\)

- Start with the two component rates; the tangent direction must come from their ratio. - Write a tangent line at a general parameter value before imposing the external-point condition. - Keep every parameter value that satisfies both the line condition and the original curve.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=t\). 2. At parameter \(t\), the point is \((2t,t^2+1)\). Requiring its tangent line to pass through \((0,-3)\) gives \(-3-(t^2+1)=t(0-2t)\). 3. This simplifies to \(t^2=4\), so \(t=-2\) or \(t=2\). 4. At \(t=-2\), the rates are \(2\) and \(-4\), the slope is \(-2\), and the tangent is \(y=-2x-3\). 5. At \(t=2\), the rates are \(2\) and \(4\), the slope is \(2\), and the tangent is \(y=2x-3\).

\(t=-2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-4\), \(\frac{dy}{dx}=-2\), \(y=-2x-3\) \(t=2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\), \(y=2x-3\)

- Convert perpendicularity into a relationship between the two line slopes. - Use the component rates to express the curve’s tangent slope in terms of \(t\). - Check that each candidate is in the interval and has a defined tangent slope.

1. The line \(4x+9y=12\) has slope \(-\frac{4}{9}\), so a perpendicular tangent must have slope \(\frac{9}{4}\). 2. The parametric tangent slope is \(\frac{dy}{dx}=\frac{3t^2-3}{2t}\), where \(t\ne0\). 3. Solving \(\frac{3t^2-3}{2t}=\frac{9}{4}\) gives \(2t^2-3t-2=0\), so \(t=-\frac{1}{2}\) or \(t=2\). 4. The corresponding points are \(\left(\frac{5}{4},\frac{11}{8}\right)\) and \((5,2)\).

\(t=-\frac{1}{2}\) at \(\left(\frac{5}{4},\frac{11}{8}\right)\) \(t=2\) at \((5,2)\)

- A tangent line must have both the correct direction and the correct point of contact. - Use the line’s slope to determine the unknown coefficient. - After finding the coefficient, check the curve’s point against the full line equation.

1. At \(t=1\), the tangent slope is \(\frac{3}{a+2}\). 2. Matching the slope of \(y=2x+1\) gives \(\frac{3}{a+2}=2\), so \(a=-\frac{1}{2}\). 3. The corresponding point is \(\left(\frac{1}{2},2\right)\). 4. Substitution into the stated line gives \(2=2\left(\frac{1}{2}\right)+1\), so the point lies on the line.

\(a=-\frac{1}{2}\); the point is \(\left(\frac{1}{2},2\right)\), which lies on \(y=2x+1\).

- Examine the tangent slope for nearby nonzero parameter values. - Simplify before taking the parameter toward the singular value. - Use the limiting direction together with the point at the singular value.

1. For \(t\ne0\), the tangent slope is \(\frac{dy}{dx}=\frac{3t^2}{2t}=\frac{3t}{2}\). 2. As \(t\to0\), the tangent slope approaches \(0\). 3. The curve passes through \((0,0)\), so the limiting tangent line is \(y=0\).

\(y=0\)

- A specified tangent line gives one condition from its slope and another from the point of contact. - Express the point at the stated parameter value using both unknown constants. - Solve the two resulting conditions together.

1. At \(t=1\), the point is \((1+a,1+b)\). Because it lies on \(y=2x-1\), \(1+b=2(1+a)-1\), so \(b=2a\). 2. The tangent slope at \(t=1\) is \(\frac{3+b}{2}\). 3. Matching the stated line’s slope gives \(\frac{3+b}{2}=2\), so \(b=1\). 4. From \(b=2a\), \(a=\frac{1}{2}\).

\(a=\frac{1}{2}\) and \(b=1\)

- Write the tangent line at a general parameter value. - Express its y-intercept in terms of the parameter. - Use the given intercept to find the point of tangency.

1. For \(t\ne0\), the tangent slope is \(\frac{3t}{2}\). 2. The tangent line at parameter \(t\) has y-intercept \(t^3+1-\frac{3t}{2}(t^2)=1-\frac{t^3}{2}\). 3. Setting the intercept equal to \(-3\) gives \(1-\frac{t^3}{2}=-3\), so \(t=2\). 4. The point is \((4,9)\), the slope is \(3\), and the tangent line is \(y=3x-3\).

The point is \((4,9)\), and the tangent line is \(y=3x-3\).

- The tangent slope must be built from the two component rates at \(t=1\). - Substitute the parametric coordinates into the tangent-line equation to locate later intersections. - A repeated parameter root corresponds to the tangency itself.

1. At \(t=1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3\), so \(\frac{dy}{dx}=\frac32\). The point is \((2,1)\). 2. The tangent line is \(y-1=\frac32(x-2)\), or \(y=\frac32x-2\). 3. A second intersection satisfies \(t^3=\frac32(2t)-2=3t-2\). 4. Thus \(t^3-3t+2=(t-1)^2(t+2)=0\). The other parameter is \(t=-2\), giving \((-4,-8)\).

At \(t=1\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3\), \(\frac{dy}{dx}=\frac32\) Tangent line: \(y=\frac32x-2\) Other intersection: \((-4,-8)\)

- Convert the direction angle into a tangent-slope condition. - Solve the resulting equation only for positive parameter values. - Evaluate both coordinate functions for each valid solution.

1. A direction angle of \(\arctan(2)\) corresponds to tangent slope \(2\). 2. The parametric tangent slope is \(\frac{3t^2+1}{2t}\). 3. Solving \(\frac{3t^2+1}{2t}=2\) gives \(3t^2-4t+1=0\), so \(t=\frac{1}{3}\) or \(t=1\). 4. The corresponding points are \(\left(\frac{10}{9},\frac{10}{27}\right)\) and \((2,2)\).

\(\left(\frac{10}{9},\frac{10}{27}\right)\) and \((2,2)\)

- Treat the horizontal-tangent case separately because its normal is vertical. - For nonzero \(t\), obtain the tangent slope from the component-rate ratio before taking a negative reciprocal. - Impose the external-point condition on each normal line.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so the tangent slope is \(t\). 2. At \(t=0\), the tangent is horizontal, so the normal is vertical. The point is \((0,-1)\), and the normal \(x=0\) passes through \((0,3)\). 3. For \(t\ne0\), the normal slope is \(-\frac1t\). Requiring the normal through \((2t,t^2-1)\) to contain \((0,3)\) gives \(3-(t^2-1)=(-\frac1t)(0-2t)\). 4. Thus \(4-t^2=2\), so \(t=\pm\sqrt2\). 5. At \(t=\sqrt2\), the tangent slope is \(\sqrt2\) and the normal is \(y-3=-\frac{x}{\sqrt2}\). At \(t=-\sqrt2\), the tangent slope is \(-\sqrt2\) and the normal is \(y-3=\frac{x}{\sqrt2}\).

\(t=0\): tangent slope \(0\), normal \(x=0\) \(t=\sqrt2\): tangent slope \(\sqrt2\), normal \(y-3=-\frac{x}{\sqrt2}\) \(t=-\sqrt2\): tangent slope \(-\sqrt2\), normal \(y-3=\frac{x}{\sqrt2}\)

- Continuity and tangent agreement impose different conditions at the joining parameter. - Compare the actual coordinate rates on the two pieces; their horizontal rates are not the same. - Solve the tangent condition before returning to the position condition.

1. The left-hand piece reaches \((2,2)\) at \(t=1\). Continuity of the right-hand piece therefore requires \(a+b=2\). 2. On the left, \(\frac{dx}{dt}=2t\) and \(\frac{dy}{dt}=3t^2\), so the tangent slope at \(t=1\) is \(\frac32\). 3. On the right, \(\frac{dx}{dt}=a\) and \(\frac{dy}{dt}=2t\), so the right-hand slope at \(t=1\) is \(\frac{2}{a}\). 4. Matching slopes gives \(\frac{2}{a}=\frac32\), so \(a=\frac43\). Then \(b=2-\frac43=\frac23\).

\(a=\frac43\) and \(b=\frac23\)

- Build the tangent slope from the component derivatives before working with the intercept. - Express the x-intercept of a general tangent in terms of \(t\). - Check every algebraic parameter solution in the original intercept condition.

1. The component rates are \(\frac{dx}{dt}=2\) and \(\frac{dy}{dt}=2t\), so \(\frac{dy}{dx}=t\). 2. For \(t\ne0\), the tangent line at \((2t,t^2+1)\) has x-intercept \(2t-\frac{t^2+1}{t}=t-\frac1t\). 3. Setting this equal to \(\frac32\) gives \(2t^2-3t-2=0\), so \(t=2\) or \(t=-\frac12\). 4. At \(t=2\), the rates are \(2\) and \(4\), the slope is \(2\), and the tangent is \(y=2x-3\). 5. At \(t=-\frac12\), the rates are \(2\) and \(-1\), the slope is \(-\frac12\), and the tangent is \(y=-\frac12x+\frac34\).

\(t=2\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4\), \(\frac{dy}{dx}=2\), \(y=2x-3\) \(t=-\frac12\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=-1\), \(\frac{dy}{dx}=-\frac12\), \(y=-\frac12x+\frac34\)

- Use the coordinate condition before working with the tangent slope. - Differentiate the two trigonometric coordinates with respect to the parameter. - Evaluate the remaining coordinate after finding both scale factors.

1. The x-coordinate condition gives \(a\cos\left(\frac{\pi}{3}\right)=3\), so \(a=6\). 2. The tangent slope is \(\frac{b\cos(t)}{-a\sin(t)}\). 3. At \(t=\frac{\pi}{3}\), the slope condition becomes \(-\frac{b}{6\sqrt{3}}=-\frac{1}{3}\), so \(b=2\sqrt{3}\). 4. The y-coordinate is \(2\sqrt{3}\sin\left(\frac{\pi}{3}\right)=3\), so the point is \((3,3)\).

\(a=6\), \(b=2\sqrt{3}\), and the point is \((3,3)\).

- Treat the two parameters independently when finding the two tangent directions. - The angle calculation uses the two tangent slopes, not the parameter values themselves. - Check that the angle formula returns the acute angle requested.

1. For \(C_1\), the component rates at \(t=0\) are \(2\) and \(1\), so its tangent slope is \(m_1=\frac12\). 2. For \(C_2\), the component rates at \(u=0\) are \(1\) and \(1\), so its tangent slope is \(m_2=1\). 3. The acute angle \(\phi\) satisfies \(\tan(\phi)=\left|\frac{m_2-m_1}{1+m_1m_2}\right|=\frac13\). 4. Therefore, \(\phi=\arctan\left(\frac13\right)\).

\(\arctan\left(\frac13\right)\)

- Compute the component-rate pair for each curve at the shared parameter value. - A common tangent requires the same point and the same tangent direction. - Decide crossing versus touching from the algebraic sign of the y-component on the two sides of \(t=0\).

1. For \(C_1\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=3t^2\), and \(\frac{dy}{dx}=\frac32t^2\), so at \(t=0\) the rates are \(2\) and \(0\) and the tangent slope is \(0\). 2. For \(C_2\), \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=4t^3\), and \(\frac{dy}{dx}=2t^3\), so at \(t=0\) the rates are \(2\) and \(0\) and the tangent slope is \(0\). 3. Both curves pass through \((0,0)\), so their common tangent line is \(y=0\). 4. For \(C_1\), \(t^3\) changes sign across \(0\), so the curve crosses the tangent line. For \(C_2\), \(t^4\ge0\) on both sides, so it only touches the line.

\(C_1\) at \(t=0\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=0\), \(\frac{dy}{dx}=0\) \(C_2\) at \(t=0\): \(\frac{dx}{dt}=2\), \(\frac{dy}{dt}=0\), \(\frac{dy}{dx}=0\) Common tangent: \(y=0\) \(C_1\) crosses the tangent; \(C_2\) only touches it.