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Calculus 1 : Calculus

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Example Questions

Example Question #11 : Rate Of Flow

The volume (in gallonws) of water in a sink after the drain is opened as a function of time can be written as: $V(t)=40e^{-0.5t}$ . What is the rate of flow out of the sink at t=2?

Possible Answers:

$\frac{20}{e}$

$\frac{80}{e}$

$\frac{-80}{e}$

$\frac{-20}{e}$

-20e

Correct answer:

$\frac{20}{e}$

Explanation:

The rate of change in volume of the sink with respect to time is given as the derivative of the volume function:

$\frac{dV}{dt}=\frac{d}{dt}(40e^{-0.5t})=-20e^{-0.5t}$

The rate of change at t=2 is then:

$\frac{dV}{dt}(2)=-20e^{-0.5(2)}=\frac{-20}{e}$

However, keep in mind that this problem asks for the flow out of the sink, so a negative change in the volume means a positive outflow. Therefore, the flow out of the sink is $\frac{20}{e}$

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Example Question #16 : Rate Of Flow

The volume of a sink with a newly open drain is a function of time, given as: $V(t)=40e^{-0.5t}$ .

Determine an equation that models the rate of change of flow into the sink.

Possible Answers:

$20e^{-0.5t}$

$-20e^{-0.5t}$

$-40e^{-0.5t}$

$10e^{-0.5t}$

$-10e^{-0.5t}$

Correct answer:

$10e^{-0.5t}$

Explanation:

The flow of volume into the sink can be found as the derivative of the volume function:

$\frac{dV}{dt}=\frac{d}{dt}(40e^{-0.5t})=-20e^{-0.5t}$

Note that as a negative value, water is flowing out of the tank. However, the problem asked for not the flow, but the rate of change of the flow, which can be found by deriving the flow function:

$\frac{d}{dt}(-20e^{-0.5t})=10e^{-0.5t}$

As the sign is opposite to that of the flow, it means that the flow slows over time.

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Example Question #11 : Rate Of Flow

A spherical balloon is being filled with water. If, at a moment in time, the balloon has a diameter of 10cm which is increasing at a rate of $0.1\frac{cm}{s}$ , what is the rate of flow of water into the balloon?

Possible Answers:

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

$2.5\frac{cm^3}{s}$

$10\frac{cm^3}{s}$

$5\frac{cm^3}{s}$

$20\frac{cm^3}{s}$

Correct answer:

$5\frac{cm^3}{s}$

Explanation:

The volume of a sphere, in terms of its diameter, is given as:

$V=\frac{\pi}{6}D^3$

Since the change in volume is equivalent to the rate of flow, we can find the latter by deriving the above equation:

$\frac{dV}{dt}=\frac{\pi}{2}D^2\frac{dD}{dt}$

The diameter is 10cm and its rate of change is $0.1\frac{cm}{s}$ .

Therefore, the rate of flow is:

$\frac{\pi}{2}(10cm)^2\left(.1\frac{cm}{s}\right)=5\frac{cm^3}{s}$

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Example Question #18 : Rate Of Flow

The volume of a fluid in tank is given by the formula $V=20-20e^{-2t}$ . What is the rate of flow into the tank? Does flow increase or decrease over time?

Possible Answers:

20e^-0.5t; increases

$40e^{-2t};increases$

$-40e^{-2t};increases$

$-40e^{-2t};decreases$

$40e^{-2t};decreases$

Correct answer:

$40e^{-2t};decreases$

Explanation:

Rate of flow can be found by taking the time derivative of the volume function:

$\frac{dV}{dt}=\frac{d}{dt}(20-20e^{-2t})=40e^{-2t}$

To determine whether the flow increases or decreases over time, take the derivative of it and check the sign:

$\frac{d}{dt}(40e^{-2t})=-80e^{-2t}$

For all t>0 , the function is negative, so flow decreases over time.

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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:

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

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

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

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

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 #21 : Rate

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:

$\$4673.10/\textup{hr}$

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

$\$326.90/\textup{hr}$

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

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 #22 : Rate

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:

27.5gallons

32.5gallons

1.3gallons

13.8gallons

2.5gallons

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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : Rate Of Flow

Example Question #16 : Rate Of Flow

Example Question #11 : Rate Of Flow

Example Question #18 : Rate Of Flow

Example Question #21 : Rate

Example Question #1811 : Functions

Example Question #23 : Rate

Example Question #1812 : Functions

Example Question #21 : Rate

Example Question #22 : Rate

All Calculus 1 Resources

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