Help pls!! (1 Viewer)

[nrumble42]

Sand is poured into a pile 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 cubic centimeters per minute. When the pile is 10cm high, how fast is:

i ) the height increasing
ii) the area of the base increasing

 

[pikachu975]

i) V = pi*r^2*h / 3
dh/dt = dh/dV * dV/dt
= pi*r^2 /3 * 4
= pi * 100 /3 * 4
=400pi/3 cm/minute

ii) Using pythagoras, L = sqrt(r^2 + h^2) where L is the slant height
A = pi*r^2 + pi*r*L
= pi*(100) + pi(10)(sqrt(100+100))
= 100pi + 10pi*10sqrt(2)
= 100pi (1+sqrt(2)) cm/minute

EDIT: I didn't even find the rate of change of surface area hahaha rip

EDIT: Area of base = A = pi*r^2
V = Ah
dA/dt = dA/dV * dV/dt
= 1/h * 4
= 4/10 = 2/5 cm^2 per minute


KEY: Chain Rule
Did this without drawing a diagram so let me know if I made a mistake.

 

[integral95]

i) V = pi*r^2*h / 3
dh/dt = dh/dV * dV/dt
= pi*r^2 /3 * 4
= pi * 100 /3 * 4
=400pi/3 cm/minute

ii) Using pythagoras, L = sqrt(r^2 + h^2) where L is the slant height
A = pi*r^2 + pi*r*L
= pi*(100) + pi(10)(sqrt(100+100))
= 100pi + 10pi*10sqrt(2)
= 100pi (1+sqrt(2)) cm/minute


KEY: Chain Rule
Did this without drawing a diagram so let me know if I made a mistake.

Aye Q2 is just asking for the change in the area of the base, not the entire surface area.