Average Rate of Change: Shampoo drips from a crack in the side of a plastic bottle...

[SimpleHero]

Problem: Shampoo drips from a crack in the side of a plastic bottle. The amount of ounces, L, that leaks is modeled by the equation L(t)=(4/3)(1+t^(3/2))^(1/2), where t is the number of minutes the leak occurs. What is the average rate the shampoo leaks during the first 9 minutes?
Do I just get the points at t= 0 and t=9 then do (y2 - y1)/(t2 - t1), no derivatives or anything?

 

[stapel]

Shampoo drips from a crack in the side of a plastic bottle. The amount of ounces, L, that leaks is modeled by the equation L(t)=(4/3)(1+t^(3/2))^(1/2), where t is the number of minutes the leak occurs. What is the average rate the shampoo leaks during the first 9 minutes?

Do I just get the points at t= 0 and t=9 then do (y2 - y1)/(t2 - t1), no derivatives or anything?

Yes; this is the "average" (as opposed to the "instantaneous") rate of change. (here)