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54245812
The volume \(V\) of liquid in a container is measured in gallons, and time \(t\) is measured in minutes. Average rates \(\frac{\Delta V}{\Delta t}\) over intervals near \(t=6\) approach \(1.8\). Which unit is appropriate for the instantaneous rate of change of volume at \(t=6\): gallons, minutes per gallon, gallons per minute, or gallons per square minute? Explain from the average-rate expression.

Hints

- Read the units of the numerator and denominator in the rate quotient. - Keep the order of the quantities the same when forming the unit. - The instantaneous rate inherits the units of the nearby average rates.

Solution

1. The numerator \(\Delta V\) is measured in gallons. 2. The denominator \(\Delta t\) is measured in minutes. 3. Therefore both the average and instantaneous rates have units of gallons per minute. 4. The instantaneous rate is \(1.8\,\text{gal/min}\).

Answer

\(1.8\,\text{gal/min}\). The rate has units of change in volume divided by change in time.
54246012
For \(f(x)=|x|\), consider the average rate of change from \(x=0\) to \(x=h\), where \(h\ne0\). For positive \(h\), the average rate is \(1\). For negative \(h\), the average rate is \(-1\). Does a single instantaneous rate of change at \(x=0\) follow from these shrinking intervals? Explain.

Hints

- Compare the behavior for positive and negative \(h\). - A single rate at the instant requires agreement from both sides. - Do not average the two one-sided values together; ask whether they converge to the same number.

Solution

1. For intervals approaching \(0\) from the right, the average rates stay at \(1\). 2. For intervals approaching \(0\) from the left, the average rates stay at \(-1\). 3. Because the two sides approach different values, there is no single two-sided instantaneous rate at \(x=0\).

Answer

No. The right-side average rates approach \(1\), while the left-side average rates approach \(-1\), so no single instantaneous rate exists at \(x=0\).
54246312
A cart's average velocities over intervals shrinking toward \(t=6\) seconds approach \(-4\,\text{ft/s}\). a) What is the cart's instantaneous velocity at \(t=6\)? b) What is its instantaneous speed at \(t=6\)? c) Explain why the sign appears in one answer but not the other.

Hints

- Use the limiting average velocity directly for part a). - Think about the relationship between speed and signed velocity. - Interpret the negative sign as direction, not as a negative amount of speed.

Solution

1. The limiting average velocity gives the instantaneous velocity, so the velocity is \(-4\,\text{ft/s}\). 2. Speed is the magnitude of velocity, so the speed is \(4\,\text{ft/s}\). 3. The negative sign records direction for velocity; speed records only magnitude.

Answer

a) \(-4\,\text{ft/s}\) b) \(4\,\text{ft/s}\) c) Velocity includes direction, while speed is nonnegative magnitude.
55591612
A particle has position \(s(t)=t^2\) feet, where \(t\) is measured in seconds. For each \(h=1\), \(0.5\), and \(0.1\), compute the average rate of change on \([2-h,2]\) and on \([2,2+h]\). Use the two patterns to estimate the instantaneous rate of change at \(t=2\).

Hints

- Compute the average-rate quotient separately for an interval ending at \(t=2\) and an interval beginning at \(t=2\). - Keep the same value of \(h\) on the left and right before comparing the two rates. - A two-sided instantaneous-rate estimate is strongest when both one-sided patterns approach the same value.

Solution

1. For \(h>0\), the left-side average rate is \(\frac{s(2)-s(2-h)}{h}=\frac{4-(2-h)^2}{h}=4-h\). 2. For \(h=1,0.5,0.1\), the left-side rates are \(3\), \(3.5\), and \(3.9\,\text{ft/s}\). 3. The right-side average rate is \(\frac{s(2+h)-s(2)}{h}=\frac{(2+h)^2-4}{h}=4+h\). 4. For \(h=1,0.5,0.1\), the right-side rates are \(5\), \(4.5\), and \(4.1\,\text{ft/s}\). 5. The left-side and right-side rates both move toward \(4\,\text{ft/s}\), so the instantaneous rate at \(t=2\) is estimated as \(4\,\text{ft/s}\).

Answer

Left-side average rates: \(3\), \(3.5\), \(3.9\,\text{ft/s}\). Right-side average rates: \(5\), \(4.5\), \(4.1\,\text{ft/s}\). Estimated instantaneous rate at \(t=2\): \(4\,\text{ft/s}\).
55591712
The graph shows a curve \(f\), the point \(A=(0,0)\), six labeled points on the curve, and secant segments from \(A\) to each labeled point. Read the coordinates of \(B\) through \(G\) from the graph. Compute the slope of each secant from \(A\). Compare the slopes as the second point approaches \(A\) from the left and from the right. What instantaneous rate of change at \(x=0\) do the secants suggest?
Figure for problem 555917

Hints

- Use the grid to read each labeled endpoint before calculating any slope. - Every secant starts at \(A=(0,0)\), so compare each point's rise with its run from the origin. - Order the left-side and right-side slopes by how close their endpoints are to \(A\).

Solution

1. The labeled points are \(B=(-2,4)\), \(C=(-1,1)\), \(D=(-0.5,0.25)\), \(E=(0.5,0.25)\), \(F=(1,1)\), and \(G=(2,4)\). 2. Since \(A=(0,0)\), the corresponding secant slopes are \(-2\), \(-1\), \(-0.5\), \(0.5\), \(1\), and \(2\). 3. From the left, the slopes \(-2,-1,-0.5\) move upward toward \(0\) as the point approaches \(A\). From the right, the slopes \(2,1,0.5\) move downward toward \(0\). 4. Both one-sided secant-slope patterns suggest an instantaneous rate of change of \(0\) at \(x=0\).

Answer

\(AB:-2\), \(AC:-1\), \(AD:-0.5\), \(AE:0.5\), \(AF:1\), \(AG:2\). Both sides suggest an instantaneous rate of change of \(0\) at \(x=0\).
55591812
A square's side length is measured in centimeters, and time \(t\) is measured in seconds. The table shows average rates of change of the square's area from \(t=1\) to \(t=1+h\). Negative \(h\) values use times before \(t=1\), and positive \(h\) values use times after \(t=1\). <table><tr><th>\(h\) (seconds)</th><td>\(-0.5\)</td><td>\(-0.1\)</td><td>\(-0.01\)</td><td>\(0.01\)</td><td>\(0.1\)</td><td>\(0.5\)</td></tr><tr><th>Average area-change rate</th><td>\(5.5\)</td><td>\(5.9\)</td><td>\(5.99\)</td><td>\(6.01\)</td><td>\(6.1\)</td><td>\(6.5\)</td></tr></table> Estimate the instantaneous rate of change of area at \(t=1\), state its units, and explain how the data from both sides support your estimate.

Hints

- Separate the rows into negative and positive \(h\) values before comparing the trends. - Use the entries with \(|h|\) closest to \(0\) as the strongest numerical evidence. - The changing quantity is area, so its rate unit contains square centimeters.

Solution

1. For negative \(h\), the average area-change rates \(5.5,5.9,5.99\) approach \(6\) as \(h\) approaches \(0\) from the left. 2. For positive \(h\), reading from farther to closer inputs, the rates \(6.5,6.1,6.01\) approach \(6\) as \(h\) approaches \(0\) from the right. 3. Because both one-sided patterns approach the same value, the instantaneous area-change rate at \(t=1\) is estimated as \(6\,\text{cm}^2/\text{s}\). 4. This means that at \(t=1\), the square's area is increasing at about \(6\) square centimeters per second.

Answer

Estimated instantaneous area-change rate: \(6\,\text{cm}^2/\text{s}\). The left-side and right-side average rates both approach \(6\), so at \(t=1\) the area is increasing at about \(6\) square centimeters per second.
55591912
Three sets of average-rate evidence are proposed for a function at \(t=4\). In each set, the interval widths shrink toward \(0\). I. From the right only, the average rates are \(8.1\), \(8.01\), and \(8.001\). II. From the left, the average rates are \(7.9\), \(7.99\), and \(7.999\); from the right, they are \(8.1\), \(8.01\), and \(8.001\). III. From the left, the average rates are \(7.1\), \(7.01\), and \(7.001\); from the right, they are \(8.9\), \(8.99\), and \(8.999\). Which set supports a two-sided instantaneous rate of change at \(t=4\)? Give the estimated rate and explain why the other two sets are insufficient.

Hints

- A two-sided instantaneous rate needs evidence from both sides of the time value. - Compare the values approached by the left-side and right-side average rates. - One-sided convergence is not enough to establish a two-sided rate.

Solution

1. Set I gives evidence only from the right. It suggests a right-hand rate near \(8\), but it gives no information from the left. 2. In Set II, the left-side rates approach \(8\) from below and the right-side rates approach \(8\) from above. This supports a two-sided instantaneous rate of \(8\). 3. In Set III, the left-side rates approach \(7\) while the right-side rates approach \(9\). Because the two sides disagree, they do not support one two-sided instantaneous rate.

Answer

Set II supports a two-sided instantaneous rate, estimated as \(8\). Set I is one-sided only, and Set III has different left- and right-side limiting rates.
54245212
A ride car's height is modeled by \(H(t)=40-5(t-3)^2\) feet, where \(t\) is measured in seconds. For each of the symmetric intervals \([2,4]\), \([2.5,3.5]\), and \([2.9,3.1]\), the average rate of change of height is \(0\,\text{ft/s}\). A student says, “Because every one of those symmetric average rates is \(0\), the car is not moving vertically anywhere near \(t=3\), and its instantaneous vertical rate at \(t=3\) must be \(0\).” a) Explain why the claim about zero vertical motion near \(t=3\) is false. b) Explain why the listed symmetric average rates alone do not establish the ordinary instantaneous rate at \(t=3\). c) For \(h>0\), compute the average vertical rate on \([3-h,3]\) and on \([3,3+h]\). Use these two expressions to determine the instantaneous vertical rate at \(t=3\).

Hints

- Separate net change over an interval from what happens at individual times inside the interval. - A two-sided instantaneous rate should be supported by compatible behavior from the left and from the right. - Write the average-rate quotient separately for \([3-h,3]\) and \([3,3+h]\).

Solution

1. Equal endpoint heights on a symmetric interval make the net height change zero, but they do not make the height constant throughout the interval. 2. A centered average rate uses values on both sides at once. Its convergence does not by itself show that the left-hand and right-hand average rates approach the same value. 3. Since \(H(3)=40\) and \(H(3-h)=40-5h^2\), the average rate on \([3-h,3]\) is \(\frac{40-(40-5h^2)}{h}=5h\,\text{ft/s}\). 4. Since \(H(3+h)=40-5h^2\), the average rate on \([3,3+h]\) is \(\frac{(40-5h^2)-40}{h}=-5h\,\text{ft/s}\). 5. As \(h\to0^+\), both one-sided average rates approach \(0\,\text{ft/s}\). Therefore the instantaneous vertical rate at \(t=3\) is \(0\,\text{ft/s}\).

Answer

a) Zero net height change over a symmetric interval does not mean the height is constant inside that interval. b) The symmetric averages alone do not establish matching left-hand and right-hand instantaneous rates. c) Left-side average rate: \(5h\,\text{ft/s}\). Right-side average rate: \(-5h\,\text{ft/s}\). Both approach \(0\,\text{ft/s}\), so the instantaneous vertical rate at \(t=3\) is \(0\,\text{ft/s}\).
54245912
For a function \(F\), the average rate of change from \(t=2\) to \(t=2+h\) simplifies to \(Q(h)=8+2h\) for every \(h\ne0\) near \(0\). Use this information to find the instantaneous rate of change at \(t=2\). Explain why substituting \(h=0\) into the original average-rate quotient would still be invalid even though the simplified expression has a value there.

Hints

- Distinguish the simplified expression from the original average-rate quotient. - Ask what happens to \(8+2h\) as \(h\) gets arbitrarily close to \(0\). - A removable algebraic hole in a quotient does not make a zero-width average rate defined.

Solution

1. As \(h\) approaches \(0\), the simplified average rate \(Q(h)=8+2h\) approaches \(8\). 2. Therefore the instantaneous rate at \(t=2\) is \(8\). 3. The original average-rate quotient has interval width \(h\) in its denominator, so \(h=0\) would create a zero-width interval and division by zero. 4. The value \(8\) comes from the limiting behavior of nearby nonzero \(h\), not from evaluating the original quotient at \(h=0\).

Answer

\(8\). The original quotient is undefined at \(h=0\); the instantaneous rate is obtained from the value approached as nonzero \(h\) tends to \(0\).
54246212
Average rates from intervals immediately to the left of \(t=7\) are \(4.7,\ 4.97,\ 4.997,\ 4.9997\). No information is given from the right of \(t=7\). A student concludes that the instantaneous rate at \(t=7\) must be \(5\). Is that conclusion guaranteed? Explain what the data do and do not establish.

Hints

- Identify which side of \(t=7\) the intervals come from. - Ask what evidence would be needed from the other side. - A two-sided instantaneous rate requires compatible limiting behavior from both directions.

Solution

1. The left-side average rates approach \(5\), so the data support a left-hand limiting rate of \(5\). 2. No right-side average-rate behavior is given. 3. Without compatible information from the right, a single two-sided instantaneous rate of \(5\) is not guaranteed.

Answer

No. The data establish a left-side limiting rate of \(5\), but they do not guarantee that the right side approaches the same value.
54246412
A chemical concentration is modeled only for \(t\ge0\) by \(C(t)=3+2t+0.5t^2\), where \(C\) is measured in milligrams per liter and \(t\) in minutes. The average rates from \(t=0\) to \(t=1\), \(0.1\), and \(0.01\) are \(2.5\), \(2.05\), and \(2.005\) milligrams per liter per minute. Estimate the initial instantaneous rate at \(t=0\). Why is right-side information appropriate here?

Hints

- Follow the average rates as the right endpoint gets closer to \(0\). - Check the stated time domain before expecting data from both sides. - At a domain endpoint, nearby behavior can come from the side that belongs to the domain.

Solution

1. The right-side average rates approach \(2\,\text{mg/L/min}\) as the interval width approaches \(0\). 2. The model is defined only for \(t\ge0\), so there are no model times to the left of \(t=0\). 3. The initial instantaneous rate is therefore \(2\,\text{mg/L/min}\).

Answer

\(2\,\text{mg/L/min}\). Right-side intervals are appropriate because \(t=0\) is the left endpoint of the model's time domain.

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