Stuck on another calculus question!?

Question

Okay, i'm doing a calculus question that has to do with someone driving a four hour drive regularly, and if s(t) represents her distance (in miles) from home t hours into the trip, the s(t) is given by s(t) = -5t^3+30t^2

We have already figured out parts A & B, how far from home she was after 1 hours, and how far apart the two cities are.

Part C is this:

How fast is she driving 1 hour into the trip? 1.5 hours into the trip?

The book gives me the solution, but I don't get it. It tells me to go through the four steps of finding a derivative, and that the answer is s'(t) = -15t^2 + 60t

My problem is, I don't know how they came up with that. I know in the steps I've done before my first step had to be to replace x with x+h (in the rule of f(x)). 

What's x in this question? Is it the t? or the s? and then, I'm supposed to say to let f(x) = something

What's the f(x) = something??

Sorry if this question is long, I'm trying to include as much detail as possible.

Scarlet Manuka · Accepted Answer

Normally we have f as a function of x. In this case we have s as a function of t, so use s everywhere you see f and t everywhere you see x.

So to find the derivative, we want
lim (h->0)  (s(t+h) - s(t)) / h
= lim (h->0) ((-5(t+h)^3 + 30(t+h)^2) - (-5t^3 + 30t^2)) / h
= lim (h->0) ((-5(t^3+3t^2h+3th^2+h^3) + 30(t^2+2th+h^2)) - (-5t^3 + 30t^2)) / h
= lim (h->0) (-5t^3-15t^2h-15th^2-5h^3 + 30t^2+60th+30h^2 + 5t^3-30t^2) / h
= lim (h->0) (-15t^2h-15th^2-5h^3 + 60th+30h^2) / h
= lim (h->0) (-15t^2 - 15th - 5h^2 + 60t + 30h)
= -15t^2 + 60t.

Mark S, JPAA · Answer

OK, the derivativ of distance traveled gives you velocity.  if s is the distance, then ds/dt is the velocity.  So you have your distance equation s(t) = - 5t^3 +30t^2 and to get velocity you have to differentiate s with respect to t.

Doing this gives ds/dt = s'(t) = -5*3t^(3-1) + 30*2*t^(2-1).

This simplifies to ds/dt = -15t^2 + 60t.

So x doesn't enter into this equation; it would if you had x be time, in which case you'd have the t's replaced by x's.  In that case you'd have s(x) as your function and you'd take the derivative ds/dx.  However in this equation x is t.

Hope this helps.

? · Answer

f(x) = 2x^three(5x^two – three)^6 If the reply should be simplified via factoring out the finest average element, the nice method of fixing this hindrance shall be take the go browsing either side first. Thus ln [f(x)] = 6 ln x + 6 ln (5x^two - three) => f'(x) / f(x) = 6/x + 6/(5x^two - three) * d/dx (5x^two - three) => f'(x) = 6 f(x) [ a million/x + 10x / (5x^two - three) ] => f'(x) = 12x^three(5x^two – three)^6 * [a million/x + 10x / (5x^two - three)].

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Stuck on another calculus question!?

3 Answers