conical problem

[dear2009]

Dear mathhelp participants,

I am trying to solve a conical problem dealing with


Sand is poured onto a growing conical pile of sand, at the constant rate of 5m3/sec. Suppose the side of the cone makes a 45 degree angle to cone’s base. How fast is the height of the sand pile increasing when the volume of the cone is 100 m3 ? Note that the answers given are approximate, in units of centimeters per second: 100cm = 1m

I got 7.60 as my answer



Thanks in advance

 

[BigGlenntheHeavy]

Your answer is correct, but it is 7.6cm./sec.

 

[dear2009]

Hey BigGlennTheHeavy,



thanks for the response. do you mind showing how you did this problem to solve it?

 

[BigGlenntheHeavy]

\(\displaystyle Given: \ V \ = \ \frac{\pi r^{2}h}{3}, \ \frac{dV}{dt} \ = \ 5m^{3}/sec., \ V \ = \ 100m^{3}, \ and \ angle \ = \ 45^{0}, \ (base \ to \ height).\)

\(\displaystyle Find \ \frac{dh}{dt}.\)

\(\displaystyle tan(45^{0}) \ = \ \frac{h}{r}, \ 1 \ = \ \frac{h}{r}, \ h \ = \ r, \ so \ V \ = \ \frac{\pi h^{3}}{3}\)

\(\displaystyle 100 \ = \ \frac{\pi h^{3}}{3} \ \implies \ h \ = \ \bigg(\frac{300}{\pi}\bigg)^{1/3}\)

\(\displaystyle Now, \ we \ are \ ready \ to \ go \ - \ ergo: \ \frac{dV}{dt} \ = \ \pi h^{2}\frac{dh}{dt}\)

\(\displaystyle 5 \ = \ \pi^{1/3}(300)^{2/3}\frac{dh}{dt}, \ \frac{dh}{dt} \ = \ .076m/sec \ = \ 7.6cm./sec \ QED\)