- #1

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- Homework Statement
- See attached

- Relevant Equations
- Related change- differentiation.

Find the problem here with solution. My interest is on part (a) only.

Solution;

The steps to solution are very much clear...the author made use of implicit differentiation. Now, would i be right to have;

##\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot \dfrac{dh}{dt}+ \dfrac{dV}{dr}\cdot \dfrac{dr}{dt}##

##-2=\dfrac{1}{3}π×\left[\dfrac{5}{14} h\right]^2\dfrac{dh}{dt}+\dfrac{2}{3}π ×\left[\dfrac{5}{14} h\right]h\dfrac{dr}{dt}##

We know that ##r=\dfrac{5}{14} h,## therefore ##\dfrac{dr}{dt}=\dfrac{5}{14}\dfrac{dh}{dt}##

##-2=\dfrac{1}{3}π×4.5918\dfrac{dh}{dt}+\dfrac{2}{3}π×12.857×\left[\dfrac{5}{14} h\right]h\dfrac{dr}{dt}##

##-2=4.8084\dfrac{dh}{dt}+9.6170\dfrac{dh}{dt}##

##-2=14.4254 \dfrac{dh}{dt}##

##\dfrac{dh}{dt}=-0.1386##

Your thoughts...i hope i did not just go round and round on the implicit approach...but hey that's how i think on this.

Solution;

The steps to solution are very much clear...the author made use of implicit differentiation. Now, would i be right to have;

##\dfrac{dV}{dt}=\dfrac{dV}{dh}\cdot \dfrac{dh}{dt}+ \dfrac{dV}{dr}\cdot \dfrac{dr}{dt}##

##-2=\dfrac{1}{3}π×\left[\dfrac{5}{14} h\right]^2\dfrac{dh}{dt}+\dfrac{2}{3}π ×\left[\dfrac{5}{14} h\right]h\dfrac{dr}{dt}##

We know that ##r=\dfrac{5}{14} h,## therefore ##\dfrac{dr}{dt}=\dfrac{5}{14}\dfrac{dh}{dt}##

##-2=\dfrac{1}{3}π×4.5918\dfrac{dh}{dt}+\dfrac{2}{3}π×12.857×\left[\dfrac{5}{14} h\right]h\dfrac{dr}{dt}##

##-2=4.8084\dfrac{dh}{dt}+9.6170\dfrac{dh}{dt}##

##-2=14.4254 \dfrac{dh}{dt}##

##\dfrac{dh}{dt}=-0.1386##

Your thoughts...i hope i did not just go round and round on the implicit approach...but hey that's how i think on this.

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