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 October 3rd, 2016, 04:57 AM #1 Member   Joined: Mar 2015 From: uk Posts: 33 Thanks: 1 Rate of change of area Can someone please help with this? A mooring buoy in the shape of a right circular cone, with the diameter of its base equal to its slant height, is submerged in the sea. Marine mud is deposited on it uniformly across the surface at a constant rate 'p'. Calculate the rate at which the surface area is increasing in terms of the height of the cone. I tried the following Area A=(pi)r^2 + (pi)rx where 'x' is the slant height so A=(pi)x^2/4 + (pi)x^2/2 = 3(pi)x^2/4 If height of cone is 'h' then h^2 = x^2 - x^2/4, so x^2 = 4h^2/3 Substituting x for h gives A=(pi)h^2 and therefore dA/dh = 2(pi)h We require dA/dt the rate of change of area using the chain rule, dA/dt = dA/dh * dh/dt Now I think that dh/dt = p/A (rate of change of volume divided by the area). This assumption may be where I'm going wrong?? So dA/dt = 2(pi)h * p/A = 2p/h But the answer in the book is 6p/h - can anyone please point out where I've gone wrong? Last edited by skipjack; October 3rd, 2016 at 09:23 AM.
October 3rd, 2016, 07:43 AM   #2
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Quote:
 Originally Posted by wirewolf Now I think that dh/dt = p/A (rate of change of volume divided by the area). This assumption may be where I'm going wrong?? So dA/dt = 2(pi)h * p/A = 2p/h But the answer in the book is 6p/h - can anyone please point out where I've gone wrong?
Yeah... this is incorrect. The statement:

"Marine mud is deposited on it uniformly across the surface at a constant rate 'p'"

seems to suggest:

$\displaystyle \frac{dV}{dt} = p$

Volume of a cone is $\displaystyle \pi r^2 \frac{h}{3}$

In our case, this becomes $\displaystyle V = \frac{\pi h^3}{9}$

Differentiate w.r.t. h to get

$\displaystyle \frac{dV}{dh} = \frac{\pi h^2}{3}$

Rearrange chain rule to get $\displaystyle \frac{dh}{dt}$

$\displaystyle \frac{dV}{dt} = \frac{dV}{dh} \cdot \frac{dh}{dt}$

$\displaystyle \frac{dh}{dt} = \frac{dV}{dt} \cdot \frac{1} {\frac{dV}{dh}} = p \cdot \frac{1}{\left(\frac{\pi h^2}{3}\right)} = \frac{3p}{\pi h^2}$

Therefore

$\displaystyle \frac{dA}{dt} = \frac{dA}{dh} \cdot \frac{dh}{dt} = 2 \pi h \cdot \frac{3p}{\pi h^2} = \frac{6p}{h}$

Last edited by skipjack; October 3rd, 2016 at 09:24 AM.

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