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Calculus 1 : How to find rate of flow

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Example Questions

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 #31 : How To Find Rate Of Flow

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:

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

$\frac{\pi}{4}$

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

$\frac{4}{\pi}$

$4\pi$

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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Example Question #37 : Rate Of Flow

A cone-shaped funnel with a base circumference of $10\pi$ and a height of is initially full of water, though water begins to drain from the tip at a rate of $32\pi$ . How fast does the water level fall when it's at a height of ?

Possible Answers:

$4\pi$

$2\pi$

$\frac{4}{\pi}$

Correct answer:

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:

$C=2\pi r$

$\frac{2\pi r}{h}=\frac{10\pi}{5}$

r=h

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

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

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

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

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

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

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=32\pi$ .

At height h=4 :

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

$32\pi=\pi 4^2 \frac{dh}{dt}$

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

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Example Question #38 : Rate Of Flow

Possible Answers:

$\frac{\pi}{5}$

$25\pi$

$\frac{5}{\pi}$

$\frac{1}{5\pi}$

$5\pi$

Correct answer:

$\frac{1}{5\pi}$

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:

d=2r

$\frac{2r}{h}=\frac{20}{2}$

r=5h

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

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

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

$V=\frac{25}{3}\pi h^3$

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

$\frac{dV}{dt}=25\pi h^2 \frac{dh}{dt}$

The term $\frac{dh}{dt}$ is unknown and is what we're solving for, but we're told that $\frac{dV}{dt}=5$ .

At height h=1 :

$5=25\pi 1^2 \frac{dh}{dt}$

$\frac{dh}{dt}=\frac{1}{5\pi}$

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Example Question #32 : How To Find Rate Of Flow

Water is being added to an empty pool using a hose that sprays out water at a constant rate of flow. The pool holds 500 gallons of water, and took 3 hours to finish filling up.

Find an expression for the total amount of water in the pool between t = 0 and t = 180 minutes.

(Hint: Find the rate in which the water was flowing out of the hose and into the pool)

Possible Answers:

278t+C

4.12t

278t

$\frac{t-500}{3}$

3.56t+C

Correct answer:

278t

Explanation:

The rate of flow for the pool can be found by taking the total amount of water that flowed from the hose (the 500 gallons) divided by the amount of time it took to give those 500 gallons (3 hours or 180 minutes). Therefore, the rate of flow of the hose is determined to be 500/180 = 2.78 gallons/min. Afterwards, in order to develop an expression for the amount of water in the pool at a given time, you can take the integral of the flow. This integral will sum all the water that has been added up to that point and give you the total amount of water in the pool.

$f(t) = \int v(t)dt = \int_{0}^{t}2.78dt= 2.78t$

This makes sense because 2.78 gallons are being added every minutes, so the total number of minutes passed multipled by the 2.78 gallons gives the total amount of gallons in the pool. Notice that the C constant was not included, this is because the pool is originally empty. If the pool was not originally empty, the intial amount of water in the pool would have been set as the value of C.

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Example Question #33 : How To Find Rate Of Flow

Determine the rate of flow of a liquid at t=5 given that it's flow function is determined by:

p(t)=-t^3+4t^2-6t+10

Possible Answers:

Correct answer:

Explanation:

To determine rate of flow, we can take the first derivative of our function by using the power rule:

p(t)=-t^3+4t^2-6t+10

p'(t)=-3t^2+8t-6

At t=5 ,

p'(5)=-3(5)^2+8*5-6=-41

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Calculus 1 : How to find rate of flow

Study concepts, example questions & explanations for Calculus 1

All Calculus 1 Resources

Example Questions

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 #31 : How To Find Rate Of Flow

Example Question #37 : Rate Of Flow

Example Question #38 : Rate Of Flow

Example Question #32 : How To Find Rate Of Flow

Example Question #33 : How To Find Rate Of Flow

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