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rates of change help (1 Viewer)

riseek

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hi people some questions from fitzpatrick:-

pg 99 Q13
sand is poured into a heap in the shape of a right circular cone whose height is always equal to the radius of the base, at a constant rate of 4 cm^3/min. when the heap is 10 cm high, how fast is the height increasing,
and the area of the base increasing.
i keep getting the wrong answer: its supposed to be 1/25pi cm/min and for second part 0.8 cm^2/ min
thanks help will be appreciated
 

shady145

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first one

so v=(1/3)pi*r^2h

and we are given the rate of change to be 4cm^3/min
so
dv/dt=4
lets call height y. so h=y
we want dy/dt.
we have dv/dt so multiplying by dy/dv should get us there
dy/dt=dv/dt * dy/dv ---------- (1)
all we need is dy/dv
v=(1/3)pi*r^2h
since the base is same as the height then r=h
then we can change the formula to
v=(1/3)pi*h^3
sub y in for h
v=(1/3)pi*y^3
dv/dy=pi*y^2
we want dy/dv, not dv/dy so just flip em
dy/dv=1/(pi*y^2)
now sub into eq (1)
dy/dt=4(1/(pi*y^2))
we want the rate when y, or h = 10, so sub in and we get
dy/dt=(1/25pi )cm/min
 

shady145

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second part
using result fomr above
dy/dt=4/(pi*y^2)
area of base=pi*r^2
since r=h=y as radius=height and height is being repped by y.
we can write it as A=pi*y^2
okay so we want dA/dt as the q asks.
dA/dt=dy/dt * dA/dy -------- (1)
we got dy/dt so lets find dA/dy, and look, A=pi*y^2 lol so
dA/dy=2pi*y
sub it into eq (1)
dA/dt=(4/(pi*y^2))x(2pi*y)
cancel
dA/dt=8/y
we want it when y=10
so
dA/dt=8/10
=0.8cm^2/min

hope u can understand my explaination
 

Aquawhite

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It's been ages since I practiced rates of change... I'd probably find it hard on first attempt too.
 

riseek

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thanks people, i understood it, juust a logical mistake i kept making, well thanks any way
 

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