MATH SOLVE

5 months ago

Q:
# This is a "water tank" calculus problem that I've been working on and I would really appreciate it if someone could look at my work and tell me if I did it properly! Thank you.

Accepted Solution

A:

Part A

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

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Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)

2pi*h^2*dh = 25*dt

int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides

(2/3)pi*h^3 = 25t + C

2pi*h^3 = 3(25t + C)

h^3 = (3(25t + C))/(2pi)

h^3 = (75t + 3C)/(2pi)

h^3 = (75t + C)/(2pi)

h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0

0 = [ (75(0) + C)/(2pi) ]^(1/3)

0 = [ (0 + C)/(2pi) ]^(1/3)

0 = [ (C)/(2pi) ]^(1/3)

0^3 = (C)/(2pi)

0 = C/(2pi)

C/(2pi) = 0

C = 0*2pi

C = 0

Therefore the h(t) function is...

h(t) = [ (75t + C)/(2pi) ]^(1/3)

h(t) = [ (75t + 0)/(2pi) ]^(1/3)

h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:

h(t) = [ (75t)/(2pi) ]^(1/3)

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Part C

Your answer is correct.

Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t

h(t) = [ (75t)/(2pi) ]^(1/3)

8 = [ (75t)/(2pi) ]^(1/3)

8^3 = (75t)/(2pi)

512 = (75t)/(2pi)

(75t)/(2pi) = 512

75t = 512*2pi

75t = 1024pi

t = 1024pi/75

At this time value, the height of the water is 8 feet

Set up the radius r(t) function

r = 2*h

r = 2*h(t)

r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t

r = 2*[ (75t)/(2pi) ]^(1/3)

dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]

dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))

dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above

dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*(1/64)

dr/dt = 25/(64pi)

getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:

r = 2h, dr/dh = 2

dh/dt = (25)/(2pi*h^2) ... from part A

dr/dt = dr/dh*dh/dt ... chain rule

dr/dt = 2*((25)/(2pi*h^2))

dr/dt = ((25)/(pi*h^2))

dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8

dr/dt = (25)/(64pi)

which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)

Everything looks good but line 4. You need to put all of the "2h" in parenthesis so the teacher will know you are squaring all of 2h. As you have it right now, you are saying "only square the h, not the 2". Be careful as silly mistakes like this will often cost you points.

============================================================

Part B

It looks like you have the right answer. Though you'll need to use parenthesis to ensure that all of "75t/(2pi)" is under the cube root. I'm assuming you made a typo or forgot to put the parenthesis.

dh/dt = (25)/(2pi*h^2)

2pi*h^2*dh = 25*dt

int[ 2pi*h^2*dh ] = int[ 25*dt ] ... applying integral to both sides

(2/3)pi*h^3 = 25t + C

2pi*h^3 = 3(25t + C)

h^3 = (3(25t + C))/(2pi)

h^3 = (75t + 3C)/(2pi)

h^3 = (75t + C)/(2pi)

h = [ (75t + C)/(2pi) ]^(1/3)

Plug in the initial conditions. If the volume is V = 0 then the height is h = 0 at time t = 0

0 = [ (75(0) + C)/(2pi) ]^(1/3)

0 = [ (0 + C)/(2pi) ]^(1/3)

0 = [ (C)/(2pi) ]^(1/3)

0^3 = (C)/(2pi)

0 = C/(2pi)

C/(2pi) = 0

C = 0*2pi

C = 0

Therefore the h(t) function is...

h(t) = [ (75t + C)/(2pi) ]^(1/3)

h(t) = [ (75t + 0)/(2pi) ]^(1/3)

h(t) = [ (75t)/(2pi) ]^(1/3)

Answer:

h(t) = [ (75t)/(2pi) ]^(1/3)

============================================================

Part C

Your answer is correct.

Below is an alternative way to find the same answer

--------------------------------------

Plug in the given height; solve for t

h(t) = [ (75t)/(2pi) ]^(1/3)

8 = [ (75t)/(2pi) ]^(1/3)

8^3 = (75t)/(2pi)

512 = (75t)/(2pi)

(75t)/(2pi) = 512

75t = 512*2pi

75t = 1024pi

t = 1024pi/75

At this time value, the height of the water is 8 feet

Set up the radius r(t) function

r = 2*h

r = 2*h(t)

r = 2*[ (75t)/(2pi) ]^(1/3) .... using the answer from part B

Differentiate that r(t) function with respect to t

r = 2*[ (75t)/(2pi) ]^(1/3)

dr/dt = 2*(1/3)*[ (75t)/(2pi) ]^(1/3-1)*d/dt[(75t)/(2pi)]

dr/dt = (2/3)*[ (75t)/(2pi) ]^(-2/3)*(75/(2pi))

dr/dt = (2/3)*(75/(2pi))*[ (75t)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

Plug in t = 1024pi/75 found earlier above

dr/dt = (25/pi)*[ (75t)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (75(1024pi/75))/(2pi) ]^(-2/3)

dr/dt = (25/pi)*[ (1024pi)/(2pi) ]^(-2/3)

dr/dt = (25/pi)*(1/64)

dr/dt = 25/(64pi)

getting the same answer as before

----------------------------

Thinking back as I finish up, your method is definitely shorter and more efficient. So I prefer your method, which is effectively this:

r = 2h, dr/dh = 2

dh/dt = (25)/(2pi*h^2) ... from part A

dr/dt = dr/dh*dh/dt ... chain rule

dr/dt = 2*((25)/(2pi*h^2))

dr/dt = ((25)/(pi*h^2))

dr/dt = ((25)/(pi*8^2)) ... plugging in h = 8

dr/dt = (25)/(64pi)

which is what you stated in your screenshot (though I added on the line dr/dt = dr/dh*dh/dt to show the chain rule in action)