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AP Calculus AB

An interactive log for students and parents in my AP Calculus class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.

Monday, April 24, 2006

Ara's Blog Assignment

An object in motion along the x-axis has velocity v(t) = (t + e^t) sin (t^2) for the interval 1, 3.

a. How many times is the object at rest?

Defining an object at rest means the slope is zero, that is, the same way as saying the derivative of the parent function is equal to zero. Therefore, since the velocity function or the first derivative is given, we only have to look for the points where v(t) is zero.

0 = (t + e^2) sin (t^2)

Plotting it into the graphing calculator, we obtain the roots at x = 1.7725 and x = 2.5066
Having these points, we say that at x = 1.7725 and x = 2.5066, the object is at rest.

b. When is the object moving to the left? Justify your answer.

A movement to the left, in my understanding, is similar to moving backward as opposed to forward. Therefore, the object is moving towards left on the interval where the velocity function is negative.

'velocity less than zero'

Looking at the graph, the points where the velocity is negative are pretty obvious. Having said that, on the interval [1.7725 , 2.5066] the object is moving to the left.

c. During the interval found in part b, when is the speed of the object increasing?

In this question, the acceleration or the second derivative is needed since the speed of the object is being asked.

a(t) = (1 + e^t) (sin (t^2)) + (t + e^t) (2t) (- cos (t^2))

*it is actually much easier if you guys just plug the function into the calculator at Y2 using nDeriv.

Finding the interval where the speed of the object is increasing also means looking for the positive values of a(t).

The roots of a(t) are located at x = 1.377 , x = 2.2161 and x = 2.8307.

Now, to the left of x = 1.377 the graph is positive and to its right negative. So on the interval 1 , 1.377 the speed is increasing.

at x = 2.2161, the graph is negative to its left and positive to its right. on the other hand, at x = 2.8307, the graph has its positive values on its left.

But, since the interval is specified, we only have to consider the roots between x = 1.7725 and x = 2.5066. Therefore, the object's speed is increasing on the interval 2.2161 , 2.5066

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