1. 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

2. Approach 1.Identify what you are trying to find. 2.Draw pictures 3.Identify variables present in problem 4.Write chain rule so you end up with the desired rate. 5.Find equations to substitute into chain rule 6.Evaluate the derivative 7.Solve problem 8.Answer in context.

3. 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 1280cm 3 /s (V=(4/3)πr 3 ) Calculate the rate at which the radius is increasing at a time 1min after the strongman has started inflating the bottle.

4. 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=πr 2 dA/dr=2πr dr/dt=2 When r=8, dA/dt=32π=100.5m 2 /s

5. A circus strongman is inflating a spherical rubber hot-water bottle at a steady rate of 1280cm 3 /s (V=(4/3)πr 3 ) 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 πr 2 dr/dt=1280/4 πr 2 When t=60, dr/dt=??? 1.Find r when t=60 (V increases by 1280cm 3 /s : after 60 sec, V=76800, use V=(4/3)πr 3 to find r=26.4cm 2.Substitute 4=26.4 into dr/dt=1280/4 πr 2 1.dr/dt=0.146cm/s after 60 seconds