A calculus rate-of-change question (1 Viewer)

c0okies

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hi could someone help me with this question pls? i kno how to do it but i cant quite get the radius in terms of the height..

Sand is poured into a heap in the shape of a right circular cone whose semi-vertex angle is alpha, where tan alpha=3/4. When the height of the cone is 16cm, the height is increasing at the rate of 2cm/min. At what rate is the volume increasing at that instant?
 
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Riviet

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Let the volume be v, height be h, radius be r, and time be t.
v = pi.r2h
dv/dh = pi.r2
dh/dt = 2

We want to find dv/dt,

dv/dt=dh/dt x dv/dh

= 2.pi.r2

That's what I've figured out so far, which angle is the semi-vertex angle referring to?
 
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c0okies

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ohhhh i think the semi vertex is the height against the edge of the cone??

actually thats what im most unsure about which is why i cant find r in terms of h
 

Riviet

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I think you mean (1) in my attachment, someone correct me if I'm wrong. ;)
 
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Riviet

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I think I've got the answer:

tan α = 3/4, which is equivalent to tan α = 12/16, where 16 is the height and 12 corresponds to the radius.

So dv/dt = dv/dh . dh/dt

= 2.π.r2

= 2.π.144

= 288π cm3/min.
 

c0okies

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=O =D it is!! thats the answer at the bottom of the page! thankQ riviet!!! xD
 

Riviet

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Yay! It wasn't that hard either! I'm glad I got it right. :)
 

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