C4 - Connected Rates of Change

I have done 11/13 questions. If anyone can help me on these i would be extremely grateful.

1) The area of a circle is decreasing at a rate of 0.5cm²/s. Find the rate of change of the circumference when the radius is 2cm.

I got dA/dt = -0.5
i need to find dC/dt

2) A hollow right circular cone has height 18cm and base radius 12cm. It is held vertex downwards beneath a tap leaking at the rate of 2cm³/s. Find the rate of the rise of water level when the depth is 6cm.

V = 1/3πr²
dV/dr = 2/3πr
dV/dt = 2

I don't know what i need to find on this one.

Thanks for all your help!

Reply 1

thefish_uk

Srathmore

I have done 11/13 questions. If anyone can help me on these i would be extremely grateful.

1) The area of a circle is decreasing at a rate of 0.5cm²/s. Find the rate of change of the circumference when the radius is 2cm.

I got dA/dt = -0.5
i need to find dC/dt
We know the following:

dA/dt = -0.5

C = 2πr
dC/dr = 2π

A = πr²
dA/dr = 2πr

We want dC/dt in terms of r

DA/dt = dA/dr * dr/dt
0.5 = 2πr * dr/dt
dr/dt = 1/(4πr)

So dC/dt = dr/dt * dC/dr
= 1/(4πr) * 2π
= 1/(2r)

So r = 2 so dC/dt = 1/4

Reply 2

thefish_uk

Srathmore

2) A hollow right circular cone has height 18cm and base radius 12cm. It is held vertex downwards beneath a tap leaking at the rate of 2cm³/s. Find the rate of the rise of water level when the depth is 6cm.

V = 1/3πr²
dV/dr = 2/3πr
dV/dt = 2

The volume of water in the cone can be expressed as V=(1/3)πr²h where h is the depth of water and r is the radius of the water surface. However we know that because this cone of water is restricted by the shape of the hollow cone it is, r:h must always be 12:18 so r=(2/3)h

So substituting into our formula, at any time, V=(1/3)π((2/3)h)²h

V = πh³(4/27)

So dV/dh = (4/9)πh²

dV/dt = dV/dh * dh/dt and we want dh/dt

2 = (4/9)πh² * dh/dt

dh/dt = 9/(2πh²)
so h=6, you can sub into that formula