Anonymous asked in Science & MathematicsMathematics · 1 month ago

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

  • Vaman
    Lv 7
    1 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

  • Pope
    Lv 7
    1 month ago


    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.


    average velocity on [2, 7]

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

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

    = -54 ms⁻¹


    I can make no grammatical sense of your sentence here.

  • Dylan
    Lv 6
    1 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

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