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Calculus 1 : Calculus

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Example Questions

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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Example Question #31 : How To Find Rate Of Flow

Water is spilling from a cone at $2cm^{3}/min$ . The cone has a radius of 16cm at the top and is 25cm high. At what rate is the depth of the water changing when the water reaches a height of 12cm ?

Possible Answers:

$\frac{dh}{dt}=\frac{25}{12\pi}$

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

$\frac{dh}{dt}=\frac{25}{16\pi}$

$\frac{dh}{dt}=\frac{625}{12288\pi}$

$\frac{dh}{dt}=\frac{1}{8\pi}$

Correct answer:

$\frac{dh}{dt}=\frac{625}{12288\pi}$

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 .

$\frac{r}{h}=\frac{16}{25}$

$r=\frac{16}{25}h$

The volume equation is now

$V=\frac{\pi }{3}\left ( \frac{16}{25}h \right )^{2}h=\frac{256\pi }{1875}h^{3}$

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.

$\frac{dV}{dt}=\frac{d}{dt}\left [\frac{256\pi }{1875}h^{3} \right ]=\frac{256\pi }{1875}*3h^{2}\frac{dh}{dt}$

$\frac{dV}{dt}=\frac{256\pi }{625}h^{2}\frac{dh}{dt}$

We are given the following parameters

$\frac{dV}{dt}=2cm^{3}/min$ and we need to find $\frac{dh}{dt}$ at h=12cm .

Subsituting this information into the first derivative

$\frac{dV}{dt}=\frac{256\pi }{625}h^{2}\frac{dh}{dt}$

$3=\frac{256\pi }{625}(12)^{2}\frac{dh}{dt}$

$3=\frac{36864\pi }{625}\frac{dh}{dt}$

$\frac{dh}{dt}=3*\frac{625}{36864\pi}=\frac{625}{12288\pi}$

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Example Question #31 : How To Find Rate Of Flow

Water is poured into a cone shaped cup at a rate of 1 m/s. The radius of the cone is 5 m. What is the rate of change of the height of the water in the cone?

Possible Answers:

$\frac{1}{9\pi}$

$\frac{1}{9}$

$\frac{1}{25\pi}$

$\frac{1}{25}$

Correct answer:

$\frac{1}{25\pi}$

Explanation:

The equation for the volume of a cone is $V=\pi r^2h$ . To find the rate of change of height we must differentiate with respect to h and treat r as a constant.

$V=(5)^2\pi h$

$\frac{dV}{dt}=25\pi \frac{dh}{dt}$

We can plug in the flow rate of 1 and solve for the rate of change of height.

$1=25\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{25\pi}$

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Example Question #31 : How To Find Rate Of Flow

A cylindrical water tank has a 2 feet radius and is 5 feet long. If the tank is halfway full of water that weighs 25 pounds per cubic feet, what is the force by the water acting on one of the tank's ends?

Possible Answers:

$\frac{400}{3}$

$\frac{800}{3}$

$\frac{200}{3}$

$\frac{50}{3}$

$\frac{500}{3}$

Correct answer:

$\frac{400}{3}$

Explanation:

For this problem, the length of the tank is irrelevant.

Write the formula for the force exerted by a fluid.

$F= \int_{a}^{b} \rho h(y)L(y)\:dy$

We will need to find the equation of the of circle given the radius.

The equation of a circle is: x^2+y^2 =r^2

The center of the circle is (0,0) .

Substitute the radius and simplify: x^2+y^2 =4

Rewrite the equation in terms of .

x^2=4-y^2

$x=\sqrt{4-y^2}$

The length of the horizontal rectangular strips are twice the length of :

$L(y) = 2\sqrt{4-y^2}$

The height h(y) , or the depth, is assuming that the downward direction of the y-axis is positive.

The pressure $\rho$ is $25\: \frac{lbs}{ft^3}$ .

The bounds are from the center of the circle to the outer rim of the tank, which is a distance of two. The bounds are from zero to two. Write the integral and solve.

$F= \int_{0}^{2} (25\: \frac{lbs}{ft^3}) (y)(2\sqrt{4-y^2})\:dy$

$F= \int_{0}^{2} 50y\sqrt{4-y^2}\:dy$

Use substitution to solve.

u=4-y^2

$du=-2y \:dy$

$-\frac{1}{2}du=y \:dy$

Replace the variables of to in terms of in order to solve the integral. Pull out all constants out in front of the integral.

$F= 50\cdot -\frac{1}{2}\int \sqrt{u}\:du = -25\times \frac{2}{3}u^{\frac{3}{2}}$

Re-substitute u=4-y^2 .

$F=-\frac{50}{3}(4-y^2)^{\frac{3}{2}}|_{0}^{2} = 0-[(-\frac{50}{3})(8)] = \frac{400}{3}$

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Example Question #34 : Rate Of Flow

A cone-shaped funnel with a base diameter of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the water level fall when it's at a height of ?

Possible Answers:

$\frac{12}{5}$

$\frac{12}{5\pi}$

$\frac{5}{12\pi}$

$\frac{5\pi}{12}$

$\frac{12\pi}{5}$

Correct answer:

$\frac{5}{12\pi}$

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone. To model this, relate the radius of the plane of the water to the height of the water:

d=2r

$\frac{2r}{h}=\frac{6}{5}$

$r=\frac{3}{5}h$

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi (\frac{3}{5}h)^2 h$

$V=\frac{3}{5}\pi h^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=\frac{9}{5}\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=3$ .

At height h=2 :

$3=\frac{9}{5}\pi 2^2 \frac{dh}{dt}$

$3=\frac{36}{5}\pi \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{5}{12\pi}$

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Example Question #1827 : Functions

A cone-shaped funnel with a base diameter of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the radius of the water's surface diminish when it's at a height of ?

Possible Answers:

$\frac{15}{196\pi}$

$\frac{3}{5\pi}$

$\frac{15\pi}{196}$

$\frac{60}{49\pi}$

$\frac{15}{98\pi}$

Correct answer:

$\frac{15}{98\pi}$

Explanation:

Itmay be guessed that the radius of the water's surface will decrease faster as it lowers towards the tip of the cone, i.e it's not constant. To model this, relate the radius of the plane of the water to the height of the water:

d=2r

$\frac{h}{2r}=\frac{10}{8}$

$h=\frac{5}{2}r$

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

$V=\frac{1}{3}\pi r^2 (\frac{5}{2}r)$

$V=\frac{5}{6}\pi r^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=\frac{5}{2}\pi r^2 \frac{dr}{dt}$

The term $\frac{dr}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=3$ .

At height h=7 :

$r=\frac{2}{5}h=\frac{14}{5}$

$\frac{dV}{dt}=\frac{5}{2}\pi r^2 \frac{dr}{dt}$

$3=\frac{5}{2}\pi (\frac{14}{5})^2 \frac{dr}{dt}$

$3=\frac{98}{5}\pi \frac{dr}{dt}$

$\frac{dr}{dt}=\frac{15}{98\pi}$

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Example Question #2851 : Calculus

A cone-shaped funnel with a base radius of and a height of is initially full of water, though water begins to drain from the tip at a rate of . How fast does the water level fall when it's at a height of 0.5 ?

Possible Answers:

$4\pi$

$\frac{4}{\pi}$

$\frac{1}{4\pi}$

$\frac{\pi}{4}$

$\frac{1}{16\pi}$

Correct answer:

$\frac{4}{\pi}$

Explanation:

It might be assumed intuitively that the water level will decrease faster as it lowers towards the tip of the cone.

Cone drain

To model this, relate the radius of the plane of the water to the height of the water:

$\frac{r}{h}=\frac{12}{4}$

r=3h

Now, since the volume of water at a point in time is approximately a cone shape, it can be written as:

$V=\frac{1}{3}\pi r^2 h$

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

$V=3\pi h^3$

To find the rate of change, take the derivative with respect to time:

$\frac{dV}{dt}=9\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=9$ .

At height h=0.5 :

$\frac{dV}{dt}=9\pi h^2 \frac{dh}{dt}$

$9=9\pi 0.5^2 \frac{dh}{dt}$

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

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Calculus 1 : Calculus

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

Example Question #1822 : Functions

Example Question #1823 : Functions

Example Question #1824 : Functions

Example Question #2851 : Calculus

Example Question #31 : How To Find Rate Of Flow

Example Question #31 : How To Find Rate Of Flow

Example Question #31 : How To Find Rate Of Flow

Example Question #34 : Rate Of Flow

Example Question #1827 : Functions

Example Question #2851 : Calculus

All Calculus 1 Resources

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