# Calculus AB Problem

1. Feb 17, 2015

### icecubebeast

1. The problem statement, all variables and given/known data
The position of a particle moving along a line is given by s(t) = 2t^3 -24t^2 + 90t + 7 for t ≥ 0. For what values of t is the speed of the particle increasing?
(a) 3 < t < 4 only
(b) t > 4 only
(c) t > 5 only
(d) 0 < t < 3 and t > 5
(e) 3 < t < 4 and t > 5
2. Relevant equations
d/dx [k*x^n] = kn*x(n-1) power rule
d/dx [f(x) +- g(x)] = f'(x) +- g'(x)
3. The attempt at a solution
s(t) = 2t^3 -24t^2 + 90t + 7
find the derivative:
s'(t) = 6t^2 - 48t + 90
find the second derivative:
s''(t) = 12t - 48

Since the problem is asking for acceleration "For what values of t is the speed of the particle increasing?", we find the point of inflection and find the intervals.
s''(t) = 0
12t - 48 = 0
12(t - 4) = 0
t=4
Intervals: (0,4] and [4,infinity)
We plug in a number for each interval:
s''(1) = 12(1) -48
=12 - 48
=-36

s''(5) = 12(5) - 48
=60 - 48
=12

The values of t>4 are when the speed is increasing.

The problem is that the solution to this problem is (e) 3 < t < 4 and t > 5 and I don't know why?

2. Feb 17, 2015

### DEvens

Hmm... Do you suppose the author of the question is thinking "magnitude of speed" rather than just speed?

A positive speed getting smaller is not speed getting bigger. This is certain. But is a negative speed getting more negative, is the speed increasing?

3. Feb 17, 2015

### icecubebeast

Is that what your teacher had in mind when he/she asked that kind of problem?

4. Feb 17, 2015

### DEvens

Oh, no. My physics teachers were perfect in every way.

5. Feb 17, 2015

### icecubebeast

How about your calculus teacher? Did he/she ask that kind of problem?

6. Feb 17, 2015

### Ray Vickson

Standard nomenclature is that speed = absolute value of velocity. Your $s'(t)$ is velocity $v(t)$, not speed. If you plot the speed $|s'(t)|$ you will see how the book's answer arises.

If you want to do it without plotting, you need to be careful about when speed is increasing; this occurs if either (a) $v(t) < 0$ is decreasing, or (b) $v(t) > 0$ is increasing. Do you see why?

7. Feb 17, 2015

### icecubebeast

So speed is different in that increasing speed depends only on the magnitude of the increase of speed, and not dependent on direction?

8. Feb 17, 2015

### Ray Vickson

No: increasing speed means increasing magnitude of velocity. If velocity is < 0 and is becoming < 0 at an even greater rate, speed increases; if velocity is > 0 and is becoming > 0 at an even greater rate, speed increases. Just think about velocity being > 0 if we are driving East and being < 0 if we are driving West, and think about what increasing speed means in both cases.

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