Related Rates of Change

Problems where there are multiple variables & expressions.

“Rate” refers to time unless otherwise stated. We link related rates using the chain rule

Approach

1. Identify what you are trying to find. 2. Draw pictures3. Identify variables present in problem4. Write chain rule so you end up with the desired

rate. 5. Find equations to substitute into chain rule6. Evaluate the derivative7. Solve problem8. Answer in context.

Eg: When a stone is dropped into a still pond of water, a circular ripple is formed. The radius of the circle increases at a rate of

2m/s. Calculate the rate at which the area of the circle is increasing when the radius is 8m.

A circus strongman is inflating a spherical rubber hot-water bottle at a steady rate of 1280cm3/s (V=(4/3)πr3) Calculate the rate at which the radius is increasing at a time 1min after the

strongman has started inflating the bottle.

Eg: When a stone is dropped into a still pond of water, a circular ripple is formed. The radius of the circle increases at a rate of

2m/s. Calculate the rate at which the area of the circle is increasing when the radius is 8m.

• We are trying to find dA/dt when r=8m • Varibles we have are r, A, t

• A=πr2 dA/dr=2πr• dr/dt=2• When r=8, dA/dt=32π=100.5m2/s

dt

dr

dr

dA

dt

dA. rr

dt

dA 422

A circus strongman is inflating a spherical rubber hot-water bottle at a steady rate of 1280cm3/s (V=(4/3)πr3) Calculate the rate at which the radius is increasing at a time 1min after the

strongman has started inflating the bottle.

• We are trying to find dr/dt when t=60 • Varibles: r, V, t

• dV/dt=1280• dV/dr=4 πr2

• dr/dt=1280/4 πr2

• When t=60, dr/dt=???1. Find r when t=60 (V increases by 1280cm3/s : after 60 sec,

V=76800, use V=(4/3)πr3 to find r=26.4cm 2. Substitute 4=26.4 into dr/dt=1280/4 πr2

1. dr/dt=0.146cm/s after 60 seconds

dt

dV

dV

dr

dt

dr.