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Calculus 1 : Rate

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Example Questions

Example Question #11 : Rate Of Flow

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation v(t)=5t^3-4t^2+9t+10 . What is the tank’s rate of flow at t=3 in liters per minute?

Possible Answers:

120

125

105

115

110

Correct answer:

120

Explanation:

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given

v(t)=5t^3-4t^2+9t+10 and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then

v'(t)=15t^2-8t+9 .

Therefore, at t=3 ,

v'(3)=15(3)^2-8(3)+9=135-24+9=120 liters per minute.

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Example Question #11 : Rate Of Flow

The volume of water a pipe recieves is given as $V(t)=\sqrt{t}+60$ . What is the flow rate of the pipe at t=16 seconds ? The volume is in liters.

Possible Answers:

$0\frac{liters}{second}$

$\frac{1}{8}\frac{liters}{second}$

$-\frac{1}{2}\frac{liters}{second}$

$-\frac{1}{8}\frac{liters}{second}$

$\frac{1}{2}\frac{liters}{second}$

Correct answer:

$\frac{1}{8}\frac{liters}{second}$

Explanation:

Flow rate can be defined as the volume of a liquid passing through a surface per time . This means that in order to solve this equation we must differentiate the volume equaiton we are given with respect to time. By doing so we will obtain the change in volume per unit time, or the flow rate. To take the derivative of this equation, we must use the power rule, $\frac{d}{dx}x^n=nx^{n-1}$ .

We also must remember that the derivative of an constant is 0. Differentiating the volume equaiton given, we obtain

$\frac{dV}{dt}=\frac{1}{2\sqrt{t}}$ .

Plugging in t=16 seconds for this equation, the flow rate of this pipe at is

$\frac{1}{8}\frac{liters}{second}$ .

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Example Question #11 : Rate

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation $v(t)=12t^{2}+10t+9$ . If Daisy were to drain the tank, what would be the rate of flow at t=2 in liters per minute?

Possible Answers:

$58\ l/m$

$78\ l/m$

$48 \ l/m$

$68\ l/m$

$38\ l/m$

Correct answer:

$58\ l/m$

Explanation:

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given

$v(t)=12t^{2}+10t+9$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then v'(t)=24t+10 .

Therefore, at t=2 ,

v'(2)=24(2)+10=48+10=58 liters per minute.

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Example Question #1805 : Functions

The volume of water (in liters) in a river at time (in minutes) is defined by the equation $v(t)=4t^{3}+5t^{2}+7$ . What is the rate of flow at t=1 in liters per minute?

Possible Answers:

$20\ l/m$

$23\ l/m$

$19\ l/m$

$22\ l/m$

$24\ l/m$

Correct answer:

$22\ l/m$

Explanation:

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given

$v(t)=4t^{3}+5t^{2}+7$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then $v'(t)=12t^{2}+10t$ .

Therefore, at t=1 ,

$v'(1)=12(1)^{2}+10(1)=12+10=22$ liters per minute.

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Example Question #1806 : Functions

The volume of water (in liters) in a water slide at time (in minutes) is defined by the equation $v(t)=7t^{3}-4t^{2}+5t-10$ . What is the rate of flow at t=3 in liters per minute?

Possible Answers:

$175\ l/m$

$155\ l/m$

$172\ l/m$

$169\ l/m$

$170\ l/m$

Correct answer:

$170\ l/m$

Explanation:

We can determine the rate of flow by taking the first derivative of the volume equation for the time provided.

Given

$v(t)=7t^{3}-4t^{2}+5t-10$ and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$ where $n\neq0$ , then $v'(t)=21t^{2}-8t+5$ .

Therefore, at t=3 ,

$v'(3)=21(3)^{2}-8(3)+5=189-24+5=170$ liters per minute.

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Example Question #11 : Rate

A spring-fed lake has a volume modeled by V(t) (in liters). Find the rate of flow after seconds and tell whether water is flowing into or out of the lake.

V(t)=16000t-5t^2

Possible Answers:

${}15,500\frac{liters}{second}$ into the lake

${}16,500\frac{liters}{second}$ out of the lake

${}15,500\frac{liters}{second}$ out of the lake

${}16,500\frac{liters}{second}$ into the lake

Correct answer:

${}15,500\frac{liters}{second}$ into the lake

Explanation:

A spring-fed lake has a volume modeled by V(t). Find the rate of flow after 50 seconds and tell whether water is flowing into or out of the lake.

To find the rate of flow from a volume function, differentiate the volume function and evaluate at the given value of t.

In other words, we need to take V(t) and find V'(5).

So, this

Becomes:

V'(t)=16000-10t

Then,

$V'(50)=16000-10(50)=15,500\frac{liters}{second}$

Since we have a positive flow rate, we can say that water is flowing into the lake, thereby increasing the volume of the lake.

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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:

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

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

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

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

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

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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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #11 : Rate Of Flow

Example Question #11 : Rate Of Flow

Example Question #11 : Rate

Example Question #1805 : Functions

Example Question #1806 : Functions

Example Question #11 : Rate

Example Question #11 : Rate Of Flow

Example Question #16 : Rate Of Flow

Example Question #11 : Rate Of Flow

Example Question #18 : Rate Of Flow

All Calculus 1 Resources

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