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 October 3rd, 2016, 05:57 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 10:23 AM. October 3rd, 2016, 08:43 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 10:24 AM. Tags area, change, rate Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post puppypower123 Calculus 1 April 14th, 2016 06:48 PM bongantedd Pre-Calculus 1 April 30th, 2014 01:40 AM joeljacks Calculus 1 October 22nd, 2012 07:32 PM Punch Calculus 2 June 6th, 2012 09:23 PM fiercedeity085 Algebra 2 March 16th, 2009 11:18 AM

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