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    A right circular cone has a constant volume. The height h and the base radius can vary. Find the rate at which h is changing with respect to r at the instant when h and r are equal.
    Ok so volume of cone is Πr^2(h/3)
    I believe the question asked for dh by dr when r=h
    So if r=h , v= Π(r^3)/3
    So dv/dr = Πr^2
    Same thing for dv/dh= Π(h^2)
    Using chain rule to get dh/dr= r^2/h^2

    Im not getting the textbook answer
    Textbook answer is -2
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    (Original post by Bilbao)
    A right circular cone has a constant volume. The height h and the base radius can vary. Find the rate at which h is changing with respect to r at the instant when h and r are equal.
    Ok so volume of cone is Πr^2(h/3)
    I believe the question asked for dh by dr when r=h
    So if r=h , v= Π(r^3)/3
    So dv/dr = Πr^2
    Same thing for dv/dh= Π(h^2)
    Using chain rule to get dh/dr= r^2/h^2

    Im not getting the textbook answer
    Textbook answer is -2
    That's because you're subbing in r=h right away without really thinking about it. The two variables change change at different rates, by making them equal before differentiation you assume they change at the same rate.

    We have V=\frac{1}{3}\pi r^2 h which is \dfrac{dV}{dr} = \frac{1}{3}\pi(2rh + r^2 \dfrac{dh}{dr} ) by the product rule + implicit differentiation.

    We have the LHS = 0 because volume remains constant, hence at r=h we get ... ?

    ALTERNATIVELY: We know V = const. so we can arrange the volume so that h= \dfrac{3V}{\pi} r^{-2} hence \dfrac{dh}{dr} = \dfrac{-6V}{\pi} r^{-3} and since the volume is \dfrac{1}{3}\pi r^3 when r=h we get the answer this way too.
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    (Original post by RDKGames)
    That's because you're subbing in r=h right away without really thinking about it. The two variables change change at different rates, by making them equal before differentiation you assume they change at the same rate.
    But aren't they changing at the same rate?
    If r increases, h will have to decrease by the same amount for V to remain constant i.e get the same value for any change in r and h?
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    Alright. Thx
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    (Original post by Carlos Nim)
    But aren't they changing at the same rate?
    If r increases, h will have to decrease by the same amount for V to remain constant i.e get the same value for any change in r and h?
    If r \mapsto 2r over a second then for the volume to remain the same we must have h \mapsto \frac{1}{4}h over the same second ie radius doubles while the height goes down to 1/4 of its size. So no, the rates at which they change are not the same. Also, they can't be because one rate must be +ve and the other must be -ve!
 
 
 
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