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Calculus 1 : Rate

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Example Questions

Example Question #21 : Rate

During a period of rainfall, the volume of fluid in a lake is given by the formula: $V(t)=15+\sqrt[3]{t^2}$

What is the rate of downpour into a lake if it is the only contributing factor to the volume's change?

Possible Answers:

$\frac{3}{2t^{\frac{2}{3}}}$

$\frac{2}{3t^{\frac{1}{2}}}$

$\frac{3}{2t^{\frac{1}{3}}}$

$\frac{2}{3t^{\frac{1}{3}}}$

Correct answer:

$\frac{2}{3t^{\frac{1}{3}}}$

Explanation:

Since the change in volume is due solely to the downpour, the rate change of the volume is equal to the rate of downpour.

For this probelm we will use the power rule to solve which states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying this rule we are able to find the rate of downpour.

$V'(t)=\frac{d}{dt}(15+\sqrt[3]{t^2})=\frac{2}{3t^{\frac{1}{3}}}$

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Example Question #1811 : Functions

Water is poured in a cylindrical container at a rate of $4 \txtup{ L/s}$ . If the radius is meters how fast is the water rising.

Possible Answers:

$9\pi$

$\frac{1}{2\pi}$

None of these

$\frac{1}{\pi}$

Correct answer:

$\frac{1}{\pi}$

Explanation:

The volume of a cylinder is $V=\pi r^{2}h$ . By implicit differentiation while holding the radius constant,

$\frac{dV}{dt}=\pi r^2\frac{dh}{dt}$

The rate of volume change is 4 and the radius is 2 so

$4=\pi*4\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{\pi}$

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Example Question #23 : Rate

Water is poured into a square pyramid shaped container at a rate of 10 L/s . The container has a side length of . What is the rate of change of the height?

Possible Answers:

None of these

Correct answer:

Explanation:

The equation for volume for the container is $V=\frac{s^{2}h}{3}$ . Implicit differentiation while holding side length constant gives

$\frac{dV}{dt}=\frac{s^2}{3}\frac{dh}{dt}$

With a side length of 3 and a flow rate of 10

$10=\frac{9}{3}\frac{dh}{dt}$

$\frac{dh}{dt}=30$

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Example Question #1812 : Functions

A cubic tank with sides of 10 feet is being filled at a rate of $Flow=10\frac{ft^3}{s}+3t^2\frac{ft^3}{s^3}$ .

If the water level in the tank is initially one foot from the bottom, how long will it tank for the tank to fill?

Possible Answers:

9.31s

10s

17.22s

5.23s

9.67s

Correct answer:

9.31s

Explanation:

Begin by finding the capacity of the tank:

$V_{max}=(10ft)^3=1000ft^2$

Now, determine the amount of water initially in the tank:

V_0=1ft(10ft)^2=100ft^3

From here, integrate the flow equation to determine the amount of water at any point in time:

$V(t)=\int Flow=\int (10\frac{ft^3}{s}+3t^2\frac{ft^3}{s^3})dt$

$V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+C$

This constant of integration can be found by using the initial condition:

$V(0)=V_0=10(0)\frac{ft^3}{s}+0^3\frac{ft^3}{s^3}+C$

C=100ft^3

$V(t)=10t\frac{ft^3}{s}+t^3\frac{ft^3}{s^3}+100ft^3$

Now, to find when the tank is full, set the equation equal to the max capacity of the tank:

$V(t_f)=10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}+100ft^3=1000ft^3$

$10t_f\frac{ft^3}{s}+t_f^3\frac{ft^3}{s^3}=900ft^3$

t_f=9.31s

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Example Question #1813 : Functions

Neo loses money such that the amount of dollars in his bank account at time (in hours) is m(t)=5000-0.0001e^t .

What is the net rate of flow of money out of Neo's bank account when t=15 ?

Possible Answers:

$-\$326.90/\textup{hr}$

$\$4673.10/\textup{hr}$

$\$326.90/\textup{hr}$

$-\$4673.10/\textup{hr}$

Correct answer:

$\$326.90/\textup{hr}$

Explanation:

The flow of money out of the bank account is the negative of the rate of change (increase) in the amount of money.

Thus, the value we are looking for is

$-m'(15)=-(-0.0001e^{15})=0.0001\cdot3268984.4=326.90$ .

So the net rate of flow out of the bank account is $326.90/hr.

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Example Question #1821 : Functions

A -gallon tank begins to drain at a rate of $0.3t^2\frac{gallon}{s^3}$ , where is a time in seconds.

How much water will remain in the tank after five seconds?

Possible Answers:

1.3gallons

13.8gallons

27.5gallons

2.5gallons

32.5gallons

Correct answer:

27.5gallons

Explanation:

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

$V(t)=\int Rdt$

$V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt$

$V(t)=-0.1t^3\frac{gallon}{s^3}+C$

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

$V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C$

C=40 gallons

Therefore the integral is:

$V(t)=40gallons -0.1t^3\frac{gallon}{s^3}$

$V(5s)=40gallons -0.1(5s)^3\frac{gallon}{s^3}$

V(5s)=27.5gallons

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Example Question #1822 : Functions

A -gallon tank begins to drain at a rate of $0.3t^2\frac{gallon}{s^3}$ , where is a time in seconds.

How much time will it take for the tank to be completely drained?

Possible Answers:

23.29s

12.67s

11.18s

7.37s

6.23s

Correct answer:

7.37s

Explanation:

To begin this problem, note that the total amount of water drained over a span of time can be found by integrating the rate function with respect to time:

$V(t)=\int Rdt$

$V(t)=\int -0.3t^2 \frac{gallon}{s^3}dt$

$V(t)=-0.1t^3\frac{gallon}{s^3}+C$

Note that the rate is treated as negative, since it is flow out of the tank. To find the value of C, the constant of integration, use the initial condition:

$V(t)=40gallons=-0.1(0^3)\frac{gallon}{s^3}+C$

C=40 gallons

Therefore the integral is:

$V(t)=40gallons -0.1t^3\frac{gallon}{s^3}$

Now to find when the tank is drained, solve for when the equation is zero:

$V(t_f)=0=40gallons-0.1t_f^3\frac{gallon}{s^3}$

t_f^3=400s^3

t_f=7.37s

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Example Question #1823 : Functions

Water flows into a cylindrical container at a rate of $3t^2\frac{ft^3}{s^3}+6t\frac{ft^3}{s^2}\textup{}$ .

If the cylinder has a radius of ten feet, a height of twenty feet, and is empty initially, how long will it take to fill completely?

Possible Answers:

9.37s

17.51s

44.29s

21.38s

10.88s

Correct answer:

17.51s

Explanation:

Begin by finding the volume of the tank:

$V_{tank}=\pi r^2h$

$V_{tank}=2000\pi ft^3$

Now, to find the amount the tank has filled over a given time, integrating the flow function with respect to time:

$V(t)=\int (3t^2\frac{ft^3}{s^3}+6t\frac{ft^3}{s^2})dt$

$V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}+C$

To find the constant of integration, use the initial condition of an empty tank:

$V(0)=0=0^3\frac{ft^3}{s^3}+3(0)^2\frac{ft^3}{s^2}+C$

C=0

$V(t)=t^3\frac{ft^3}{s^3}+3t^2\frac{ft^3}{s^2}$

Now, to find the time when the tank is full, set the equation equal to the max volume of the tank:

$V(t_f)=t_f^3\frac{ft^3}{s^3}+3t_f^2\frac{ft^3}{s^2}=V_{tank}=2000\pi ft^3$

t_f=17.51s

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Example Question #1824 : Functions

The equation for the volume of fluid over time is given by

$Q=\frac{1}{3}t^3+2.5t^2+4t+7$ , where is measured in m^3 and is measured in seconds .

Determine the rate of flow at t=4s .

Possible Answers:

$40 \frac{m^3}{s}$

$-40 \frac{m^3}{s}$

$32 \frac{m^3}{s}$

$200\frac{m^3}{s}$

Correct answer:

$40 \frac{m^3}{s}$

Explanation:

Since we're given volume and need to calculate flow, we take the derivative of volume with respect to time .

$Q=\frac{1}{3}t^3+2.5t^2+4t+7$

To find the derivative of Q{} with respect to t{} , we need to use the power rule.

Recall that the power rule for differentiation is given as

$\frac{\mathrm{d} }{\mathrm{d} t} (At^n)=nAt^{n-1}{}$ , where A,n{} are constants and t{} is a variable.

Using this rule for

$Q=\frac{1}{3}t^3+2.5t^2+4t+7{}$

$\frac{dQ}{dt}=t^2+5t+4$

At t=4,{}

$\frac{dQ(4)}{dt}=4^2+5*4+4= 40 \frac{m^3}{s}{}$ .

We know the units are in $\frac{m^3}{s}{}$ because we determined the rate of flow, which is volume of fluid moving per unit time.

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Example Question #2851 : Calculus

Water is passing through a cone-shaped filter at $1cm^{3}/min$ . The cone has a radius of 4cm at the top and is 8cm high. At what rate is the depth of the water changing when the water reaches a height of 3cm ?

Possible Answers:

$\frac{dh}{dt}=\frac{475 }{921\pi}$

$\frac{dh}{dt}=\frac{4 }{9\pi}$

$\frac{dh}{dt}=\frac{3}{8\pi}$

$\frac{dh}{dt}=\frac{4 }{8\pi}$

$\frac{dh}{dt}=\pi$

Correct answer:

$\frac{dh}{dt}=\frac{4 }{9\pi}$

Explanation:

We will use the equation for the volume of a cone and its derivative to solve this problem.

The equation for the volume of a cone is

$V=\frac{\pi }{3}r^{2}h$

where is the radius of the cone at the top and is the height of the cone.

Since we do not know $\frac{dr}{dt}$ , we must eliminate from the volume equation before we differentiate.

We use the ratio of height to radius to eliminate .

$\frac{r}{h}=\frac{4}{8}=\frac{1}{2}$

$r=\frac{1}{2}h$

The volume equation is now

$V=\frac{\pi }{3}\left (\frac{1}{2}h \right )^{2}h=\frac{\pi }{12}h^{3}$

We will now take the derivative of the equation with respect to time or . We will use the power and chain rules to find the derivative.

$\frac{dV}{dt}=\frac{d}{dt}\left [\frac{\pi }{12}h^{3}\right ]=\frac{\pi }{12}*3h^{2}\frac{dh}{dt}$

$\frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}$

We are given the following parameters

$\frac{dV}{dt}=1cm^{3}/min$ and we need to find $\frac{dh}{dt}$ at h=3cm .

Subsituting this information into the first derivative

$\frac{dV}{dt}=\frac{\pi }{4}h^{2}\frac{dh}{dt}$

$1=\frac{\pi }{4}(3)^{2}\frac{dh}{dt}$

$1=\frac{9\pi }{4}\frac{dh}{dt}$

$\frac{dh}{dt}=\frac{4 }{9\pi}$

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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #21 : Rate

Example Question #1811 : Functions

Example Question #23 : Rate

Example Question #1812 : Functions

Example Question #1813 : Functions

Example Question #1821 : Functions

Example Question #1822 : Functions

Example Question #1823 : Functions

Example Question #1824 : Functions

Example Question #2851 : Calculus

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