Isosceles Triangular Prism Related Rates Problem

maladroit
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Homework Statement



A trough is 9 ft long and its ends have the shape of isosceles triangles that are 5 ft across at the top and have a height of 1 ft. If the trough is being filled with water at a rate of 14 ft3/min, how fast is the water level rising when the water is 7 inches deep?

I know b, h, l, dv/dt, dl/dt.
I need to first find db/dt then solve for dh/dt

Homework Equations



Volume of Iso. triangular prism= 1/2bh*l
dv/dt=1/2bh(dl/dt)+l(1/2b(dh/dt)+1/2h(db/dt)

I assume that dl/dt=0, so the new equation for the derivative is equal to.. dv/dt=l(1/2b(dh/dt)+1/2h(db/dt)

The Attempt at a Solution


If anyone could just give me some direction where to start in solving for db/dt that would be great!
 
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Hi maladroit! Welcome to PF! :smile:
maladroit said:
Volume of Iso. triangular prism= 1/2bh*l
dv/dt=1/2bh(dl/dt)+l(1/2b(dh/dt)+1/2h(db/dt)

oooh, that's far too complicated! :cry:

There's a relation between b and h, so write b in terms of h, and then differentiate! :wink:
 
That seems to be just what my problem is... I can't see the relationship between b and h.

I tried setting up a triangle and solving for what b is equal to in terms of h but I am still getting the wrong answer.
I used cos(theta)=(1/2b)/h, differentiated, and solved for what db/dt was equal to.
 
maladroit said:
That seems to be just what my problem is... I can't see the relationship between b and h.

uhh? :confused: … it's a triangle … it's the same all the way up … b = 5h.
 
Once I replaced the db/dt with the 5dh/dt and solved for the problem I got .4817 ft/min but that was incorrect.

The equation I used was...
14=9(1/2*5(dh/dt)+1/2*(7/12)5(dh/dt)
 
maladroit said:
The equation I used was...
14=9(1/2*5(dh/dt)+1/2*(7/12)5(dh/dt)

Sorry, not following that :confused:

can you write it out properly: what is v in terms of h, then what is dv/dt in terms of h, then what is dv/dt when h = 7/12. :smile:
 
Right!
I used the same equation I put before
dv/dt=l(1/2b(dh/dt)+1/2h(db/dt))
and, differentiating b=5h,

dv/dt=l(1/2b(dh/dt)+1/2h(5dh/dt))

Also, I really appreciate your help!
 
hmm … v = (22.5)h2, dv/dt = … ? :wink:
 
oh! I differentiated much too early... thank you so much for your help!
 

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