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# Still need help|||Use the difference quotient to determine the average velocity

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I understand how do everything, but im getting a different answer? Dont i just plug in the values for T and H?

Veteran  Apr 15, 2017
edited by Veteran  Apr 15, 2017
edited by Veteran  Apr 15, 2017
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What is the full question?

Alan  Apr 15, 2017
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Veteran  Apr 15, 2017
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In the following h is the time increment:

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Alan  Apr 16, 2017
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Thanks Heureka,

I have answered this before I saw your answer.  I'd like to upload mine even though it is essentially the same as yours.

$$s(t)=122+45t-16t^2$$

s is height (feet),

which is a funtion of t which is time (seconds)

To derive the given answer, h is a difference in time.  Using h for this seems really confusing to me, and I suspect that this is part of your problem.

As you already know height at time t is given by

$$s(t)=122+45t-16t^2$$

h seconds later the height is given by

$$s(t+h)\\ =122+45(t+h)-16(t+h)^2\\ =122+45t+45h-16(t^2+h^2+2th)\\ =122+45t+45h-16t^2-16h^2-32th\\$$

Now the difference quotient is the  gradient of the secant joining those two elevations with regards to time.

That is

$$difference\;\; quotient \\ =\frac{difference \;in\; height}{difference \;in\;time}\\ =\frac{s(t+h)-s(t)}{(t+h)-t}\\ =\frac{(122+45t+45h-16t^2-16h^2-32th)-(122+45t-16t^2)}{h}\\ =\frac{45h-16h^2-32th}{h}\\ =45-16h-32t\\$$

So the difference quotient over the time interval   [2.1,8]  is

$$t=2.1\\ h=8-2.1=5.9\\ DQ=45-16*5.9-32*2.1 = -116.6$$

Well  Veteren, it took a while but now you have 2 solid answers  :))

Melody  Apr 16, 2017