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  • 2 Post By skeeter
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May 12th, 2017, 01:24 PM   #1
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I am struggling with this related rates problem.

bungee jump diagram.pdf

I have attached a diagram for this problem.

(ex) A bungee jumper has reached a point in her exciting plunge where the taut cord is 100 ft long with 0.5 inch radius and stretchings. She is still 80 ft above the ground and is now falling at 40 ft/sec. You are observing her jump from a spot on the ground 60 ft from the potential point of impact as shown in the diagram.

(a) Assuming the cord to be a cylinder with volume remaining constant as the as the cord stretches. At what rate is its radius changing?

(b) From your observation point, at what rate is the angle of elevation to the jumper changing

my attempt

(a) V = pi*r^2*h

Differentiating it gives 0=2r*dr/dt + r^2*dh/dt

With r=1/24 and dh/dt=40, solving the above equation gives dr/dt=5/6 ft/sec.

But, the answer key says 2 ft/sec.

(b) let y be the distance between the jumper and the ground

tan(theta) =y/60

Differentiating it and isolating d(theta)/dt give d(theta)/dt=(cos(theta))^2/60*dy/dt ------- (*)

When y=80 ft, tan(theta)=80/60. So, (cos(theta))^2=9/25.

Now, plug 9/25 into (*) above. I have d(theta)/dt=3/500 *dy/dt.

I am stuck because I don't have information about dy/dt.

Can someone please explain how to do this problem? Thank you very much for your help.
davedave is offline  
May 12th, 2017, 02:01 PM   #2
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(a) I disagree with $2 \, ft/sec$ for the radius rate of change when $r = \dfrac{1}{24} \, ft$, $h = 100 \, ft$ and $\dfrac{dh}{dt} = 40 \, ft/sec$ ...

$V = \pi r^2 h$

$\dfrac{dV}{dt} = \pi \left(r^2 \cdot \dfrac{dh}{dt} + h \cdot 2r \cdot \dfrac{dr}{dt} \right)$

$0 = r^2 \cdot \dfrac{dh}{dt} + 2rh \cdot \dfrac{dr}{dt}$

$-\dfrac{r \cdot \dfrac{dh}{dt}}{2h} = \dfrac{dr}{dt}$

$\dfrac{dr}{dt} = - \dfrac{1}{120} \, ft/sec = -\dfrac{1}{10} \, in/sec$

(b) $\theta = \arctan\left(\dfrac{y}{60}\right)$

$\dfrac{d\theta}{dt} = \dfrac{\dfrac{1}{60} \cdot \dfrac{dy}{dt}}{1 + \left(\dfrac{y}{60}\right)^2}$

at $y = 80 \, ft$, $\dfrac{dy}{dt} = -40 \, ft/sec$

$\dfrac{d\theta}{dt} = -\dfrac{6}{25} \, rad/sec$
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skeeter is offline  
May 12th, 2017, 02:27 PM   #3
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I agree with skeeter; I ran the numbers as well and $\displaystyle -\frac{1}{10}\frac{inches}{sec}$ is what I got as well.

In your formulation of $\displaystyle \frac{dV}{dt}$you forgot to include h after the 2r. See skeeter's dV/dt formula.

Last edited by skipjack; May 12th, 2017 at 04:13 PM.
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