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# Help with the rate of change question?

A function is modeled by h(t)= 108t- 18t^2 where h(t) is heigh in m and t in seconds.

a) Find the average of change on the interval [a,b]

b) Find the average velocity for the time period beginning at t=2 seconds and lasting 5 seconds.

c) Find the rate of increase of height as time c seconds.

### 3 Answers

- VamanLv 71 month ago
First integrate h(t) between b and,

Int of h(t)=f(t) will be = 54t^2-6t^3. f(b)=54 b^2-6b^3, f(a)=54 a^2-6a^3. f(b)-f(a)= (54(b^2-a^2) - 6(b^3-a^3)). Average will be (54(b^2-a^2) - 6(b^3-a^3))/(b-a)= 54(b+a) -6(b^2+ab+a^^2)

Average velocity= 54(5+2)-6(25+10+4)=378-6*39=27.

dh/dt= rate of increase in height= 108-36*c. This is the answer

- PopeLv 71 month ago
(a)

change in h(t) on interval [a, b]

= h(b) - h(a)

= 108b - 18b² - 108a + 18a²

That is the change on the given interval. As for the average change, perhaps you should explain what you mean by that.

(b)

average velocity on [2, 7]

= [h(7) - h(2)] / (7 - 2)

= [(-126) - (144)] / 5

= -54 ms⁻¹

(c)

I can make no grammatical sense of your sentence here.

- DylanLv 61 month ago
Average change is just (h(b)-h(a))/(b-a)

(h(5)-h(2))/(5-2)... Sure you can just sub in and solve.

Just take derivative of h(t)... 108 - 36c