How can I use definite integrals to solve the given problem?

Question

A particle P moves in a straight line so that at t seconds, its displacement, s meters from the starting point is s=t^3-7t^2+17t. Determine the interval of time when the particle is moving at an increasing speed.

Answer 1

This problem does not need integration to solve because the derivative of position is velocity, and the derivative of velocity is acceleration. 

What the problem is asking is what interval of time is the particle moving at increasing speed – that is, the acceleration is positive, or the curve is concave up.

If s(t) = t^3 – 7t^2 + 17t. then first derivative (velocity)

S’(t) = 3t^2-14t + 17

And acceleration is one more derivative

A(t) = 6t-14. Considering that time must be positive or >=0, and the point of inflection is determined by setting the second derivative = 0, or t = 2.3 seconds. After that time a(t)>0 so velocity is increasing at increasing rate.

Answer 2

For displacement questions:

The function, s = t³ - 7t² + 17t, gives the position of the moving particle at t seconds.

The derivative of s with respect to t, gives the velocity of the moving particle at t seconds. This function is:

s' =ds/dt = v = 3t² -14t + 17

The derivative of v with respect to t, gives the acceleration of the moving particle at t seconds. This function is:

s'' = v' = dv/dt = a = 6t - 14

We have three functions to graph in the same plane, position, velocity, and acceleration.

s = t³ - 7t² + 17t

v = 3t² - 14t + 17

a = 6t -14

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If you look at the graph of a derivative, say y', wherever y' is negative, as in below the x-axis, then y is decreasing. Wherever y' is positive (above the x-axis), then y is increasing. At the value y' crosses the x-axis (y' roots or zeros), y is at its max or minimum value, depending on whether y' goes from negative to positive (minimum) or from positive to negative (maximum).

The problem asks us to determine the interval of time at which the particle is moving with increasing speed.

The particle moving with increasing speed is the same as saying its velocity is increasing. If we want to know where the velocity is increasing, we will look at the acceleration:

1. Anywhere the acceleration values are negative, the velocity is decreasing.
2. Anywhere the acceleration value(s) = 0, the velocity is at a maximum or minimum
3. Anywhere the acceleration values are positive, the velocity is increasing

Using the graph, we can easily tell where the values for acceleration are positive:

The graph for the acceleration function shows that acceleration is linear and acceleration is zero at t=2.33 seconds. For values of t greater than 2.33, acceleration is positive.

Since wherever acceleration is positive shows where the velocity is increasing, we have our interval for when the velocity is increasing and that is when t > 2.33 or (2.33,∞).

The interval of time at which the particle is moving with increasing speed is (2.33,∞).

Wow! I didn't see the part about using definite integrals to solve this problem. You wouldn't integrate the s function and get any useful information. Maybe you take the derivatives like I did and set them equal to zero and that would give you intervals. It seems redundant though, to take the derivatives, then integrate them, but I think you must, because that may be the only way to determine your intervals.

So when you set the acceleration function to zero, you get t = 2.33 seconds. This means you have an interval from (0,2.33) and another interval from (2.33,∞). Then I suppose, you can integrate the acceleration function with those intervals??

Given the problem, I would ask why would we want to use definite integrals to solve, when we can use derivatives to solve instead.