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Calculus 1 : Calculus

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Example Questions

Example Question #1 : How To Find Rate Of Flow

The volume of water (in liters) in a tank at time (in minutes) is defined by the equation v(t)=4t^3+7t^2+10t . What is the tank’s rate of flow at t=1 in liters per minute?

Possible Answers:

Correct answer:

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+7t^2+10t , then using the power rule which states,

$f(x)=x^n\rightarrow f'(x)=nx^{n-1}$ thus v'(t)=12t^2+14t+10 .

Therefore, at t=1 ,

v'(1)=12(1)^2+14(1)+10=12+14+10=36 liters per minute.

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Example Question #2 : How To Find Rate Of Flow

The area bouned by the curve y=x^2 and y=1 is being filled up with water at a rate of unit-squared per second. When the water-level is at y=0.5 how quickly is the water-level rising?

Canvas

Possible Answers:

units per second

$\frac{\sqrt{2}}{2}$ units per second

$2\sqrt{2}$ units per second

$\sqrt{2}$ units per second

$\sqrt{\sqrt{2}}$ units per second

Correct answer:

$\sqrt{2}$ units per second

Explanation:

If y=0.5 , then $x=\pm\sqrt{2}/2$ . Therefore the width of the water at this level is $\sqrt{2}$ . Some small change in height is associated with a change of $\sqrt{2}dh$ in area. In other words:

$\frac{dA}{dh}=\sqrt{2}$

Taking the reciprocal yields:

$\frac{dh}{dA}=\frac{\sqrt{2}}{2}$

We also know that is increasing at 2 unit-squared per second, so:

$\frac{dA}{dt}=2$

Therefore:

$\frac{dh}{dt}=\frac{dh}{dA}\frac{dA}{dt}=\left(\frac{\sqrt{2}}{2}\right)\left(2\right)=\sqrt{2}$

Therefore, the height will be rising at $\sqrt{2}$ units per second.

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Example Question #3 : How To Find Rate Of Flow

The volume of water (in liters) in a stream at time (in minutes) is defined by the equation v(t)=7t^3+5t^2-10t+6 . What is the stream’s rate of flow at t=3 in liters per minute?

Possible Answers:

219

209

190

109

210

Correct answer:

209

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+5t^2-10t+6 and the power rule

$\frac{d}{dt}t^{n}=nt^{n-1}$

where $n\neq0$ , then

v'(t)=21t^2+10t-10 .

Therefore, at t=3 ,

v'(3)=21(3)^2+10(3)-10=189+30-10=209 liters per minute.

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Example Question #1 : How To Find Rate Of Flow

The volume of water (in liters) in a pool at time (in minutes) is defined by the equation v(t)=6t^3-4t^2-8t+9 . What is the pool’s rate of flow at t=4 in liters per minute?

Possible Answers:

218

258

238

288

248

Correct answer:

248

Explanation:

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

Given

v(t)=6t^3-4t^2-8t+9 and the power rule

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

v'(t)=18t^2-8t-8 .

Therefore, at t=4 ,

v'(t)=18(4)^2-8(4)-8=288-32-8=248 liters per minute.

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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:

110

115

105

120

125

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:

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

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

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

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

$0\frac{liters}{second}$

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 #1801 : Functions

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:

$68\ l/m$

$48 \ l/m$

$38\ l/m$

$58\ l/m$

$78\ l/m$

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 #12 : Rate Of Flow

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:

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

${}15,500\frac{liters}{second}$ into 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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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : How To Find Rate Of Flow

Example Question #2 : How To Find Rate Of Flow

Example Question #3 : How To Find Rate Of Flow

Example Question #1 : How To Find Rate Of Flow

Example Question #11 : Rate Of Flow

Example Question #11 : Rate Of Flow

Example Question #1801 : Functions

Example Question #1805 : Functions

Example Question #1806 : Functions

Example Question #12 : Rate Of Flow

All Calculus 1 Resources

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