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# another rates of change question watch

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1. a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

show when the depth of the water = h, V= (pi x h^3) / 12

dv/dt = 4

dh/dt = dv/dt x dh/dv

dh/dv = 2/3 x pi x r x h

I said r = h/2
and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

anyone able to show me where i went wrong
thanks
2. (Original post by hatsuko)
a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

show when the depth of the water = h, V= (pi x h^3) / 12

dv/dt = 4

dh/dt = dv/dt x dh/dv

dh/dv = 2/3 x pi x r x h

I said r = h/2
and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

anyone able to show me where i went wrong
thanks
You are still making exactly the same error as in the previous question

3. (Original post by hatsuko)
a hollow right circular cone whose height is twice the radius is held with the vertex downwards and water is poured in at the rate of 4 cm^3s^-1

show when the depth of the water = h, V= (pi x h^3) / 12

dv/dt = 4

dh/dt = dv/dt x dh/dv

dh/dv = 2/3 x pi x r x h

I said r = h/2
and ended up with (8 x pi x h^2) / 6 so ive obviously gone wrong

anyone able to show me where i went wrong
thanks
So what is V?

You need 'h' and no 'r'. So what could you replace 'r' with?

So what is V?

You need 'h' and no 'r'. So what could you replace 'r' with?
h/2 ?
5. (Original post by hatsuko)
h/2 ?
Yes.
Yes.
i subd that in and differentiated and got the answer, but why dont you have to time by 4?
7. (Original post by hatsuko)
i subd that in and differentiated and got the answer, but why dont you have to time by 4?
Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

Remember to square both the numerator and denominator.
Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

Remember to square both the numerator and denominator.
oh i see what u mean, i thought you had to x4 because water is going in ar 4 cm^3s^-1, but its just asking for the volume at h
Find (h/2)^2 and substitute it back into the equation for the volume of the cone.

Remember to square both the numerator and denominator.
there is a second part, find in terms of pi, the rate of rise of the water when the depth is 2cm I got

dh/dt = dv/dt x dh/dv

dh/dv = pih^3 / 12

h=2

answer = 8pi/3, does this look right?
10. (Original post by hatsuko)
there is a second part, find in terms of pi, the rate of rise of the water when the depth is 2cm I got

dh/dt = dv/dt x dh/dv

dh/dv = pih^3 / 12

h=2

answer = 8pi/3, does this look right?
11. (Original post by hatsuko)
...
I have to agree with TenOfThem here.

There's no point using the chain rule when you don't understand the basic rules of differentiation.

I have to agree with TenOfThem here.

There's no point using the chain rule when you don't understand the basic rules of differentiation.

oh, its wrong then?

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