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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.
54245212
A ride car's height is modeled by \(H(t)=40-5(t-3)^2\) feet. 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 average rates is \(0\), the car is not moving vertically anywhere near \(t=3\).” Evaluate the student's reasoning and state what the shrinking-interval averages do support about the instant \(t=3\).

Hints

- An average rate uses only the changes between the interval endpoints. - Equal heights at two times do not imply equal height at every time between them. - Separate the claim about a single instant from the claim about an entire neighborhood.

Solution

1. Each symmetric interval has equal endpoint heights, so its average rate of change is \(0\). 2. A zero average rate over such an interval does not imply the height is constant throughout the interval; the car can rise before \(t=3\) and fall after \(t=3\). 3. As the intervals shrink around \(t=3\), their average rates remain \(0\), supporting an instantaneous vertical rate of \(0\,\text{ft/s}\) at \(t=3\).

Answer

The student's conclusion about the entire neighborhood is false. The data support an instantaneous vertical rate of \(0\,\text{ft/s}\) at \(t=3\), not zero vertical motion at every nearby time.
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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