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Calculus 1 : Rate

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Example Questions

Example Question #1 : Rate

The volume (v) in liters of water in a tank at time in seconds is $v = 600-4\sqrt{t}$ . What is the rate of flow from the tank at t=16

Possible Answers:

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

liters per second

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

Correct answer:

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

Explanation:

To find the rate of flow, you need to differentiate the function. This is $v^{'}=\frac{-2}{\sqrt{t}}$ substituting in for gives the answser of -0.5 .

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Example Question #1 : Rate

Suppose a fish tank has a shape of a square prism with a length of 10 inches. If a hose filled the tank at 3 cubic inches per second, how fast is the water's surface rising?

Possible Answers:

$\frac{3}{25}$

$\frac{3}{50}$

$\frac{3}{100}$

$\frac{3}{1000}$

$\frac{3}{75}$

Correct answer:

$\frac{3}{100}$

Explanation:

Given the length of the cube is 10 inches, the length and the width is also 10 inches. However, the height of the water is unknown. Let's assume this height is .

Write the volume of the water in terms of .

V=10(10)y= 100y

Differentiate the volume equation with respect to time.

$\frac{dV}{dt}=100 \cdot \frac{dy}{dt}$

Substitute the rate of the water flow into $\frac{dV}{dt}$ .

$3=100 \cdot \frac{dy}{dt}$

$\frac{dy}{dt}=\frac{3}{100}$

The water is rising at a rate of $\frac{3}{100}$ inches per second.

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Example Question #2 : Rate

Make sure you identify what this question is asking.

A large vat contains 1000 L of butter. The vat has a small leak, out of which, 100 mL (0.1 L) of butter escapes every hour. What is the rate of change in the volume of butter in the vat?

Possible Answers:

$100\ mL/hr$

$0\ mL/hr$

$-100\ mL/hr$

$-0.01\ mL/hr$

$100\ L/hr$

Correct answer:

$-100\ mL/hr$

Explanation:

This question tells you that there is a leak of 100 mL/hour, and then asks you to identify how quickly the leak is causing butter to be lost.

The key is to identify the units, mL/hour, see that there are 100 mL/hour being moved, and recognize that the units are negative, as 100 mL are being removed every hour.

Thus the rate of change in the volume of the butter is $-100\ mL/hr$ .

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Example Question #3 : Rate

A cylindrical tank with a radius of 20 centimeters and an arbitrary height is filling with water at a rate of 1.5 liters per second. What is the rate of change of the water level (its height)?

Possible Answers:

$\frac{15\pi}{4 }\frac{cm}{s}$

$\frac{15\pi }{8}\frac{cm}{s}$

$\frac{15}{8\pi }\frac{cm}{s}$

$\frac{15}{4\pi }\frac{cm}{s}$

Correct answer:

$\frac{15}{4\pi }\frac{cm}{s}$

Explanation:

The first step we may take is to write out the formula for the volume of a cylinder:

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

Where r represents the radius and h the height.

With this known we can find the rate change of volume with respect to height, by deriving these functions with respect to height:

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

Since we're interested in the rate change of height, the dh term, let's isolate that on one side of the equation:

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

dV, the rate change of volume, is given to us as 1.5 liters per second, or 1500 cm³/second. Plugging in our known values, we can thus solve for dh:

$dh = \left( \frac{1500\frac{cm^{3}}{s}}{\pi (20cm)^{2}}\right) = \frac{15}{4\pi }\frac{cm}{s}$

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Example Question #4 : Rate

The volume of water (in liters) in a pool at time (in minutes) is defined by the equation v(t)=7t^2+5t-12. If Paul were to siphon water from the pool using an industrial-strength hose, what would be the rate of flow at t=2 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)=7t^2+5t-12 , we can apply the power rule,

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

Then v'(t)=14t+5 .

Therefore, at t=2 ,

v'(2)=14(2)+5=28+5=33 liters per minute.

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Example Question #5 : Rate

The volume of water (in liters) in a river at time (in minutes) is defined by the equation v(t)=8t^3+9t^2-20t+6 . What is the river’s rate of flow at t=3 in liters per minute?

Possible Answers:

150

225

300

250

Correct answer:

250

Explanation:

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

Given,

v(t)=8t^3+9t^2-20t+6 , we can apply the power rule,

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

v'(t)=24t^2+18t-20 .

Therefore, at t=3 ,

v'(3)=24(3)^2+18(3)-20=216+54-20=250 liters per minute.

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Example Question #6 : Rate

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 #4 : Rate

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:

$2\sqrt{2}$ units per second

units per second

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

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

$\sqrt{2}$ 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 #7 : Rate

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:

210

209

109

219

190

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 #8 : Rate

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:

238

218

288

248

258

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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Calculus 1 : Rate

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1 : Rate

Example Question #1 : Rate

Example Question #2 : Rate

Example Question #3 : Rate

Example Question #4 : Rate

Example Question #5 : Rate

Example Question #6 : Rate

Example Question #4 : Rate

Example Question #7 : Rate

Example Question #8 : Rate

All Calculus 1 Resources

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