Chapter 2.7, Problem 44E

To determine

To identify: Each curves on the given graph and give proper explanation.

Answer to Problem 44E

In the given graph, d is position curve, c is velocity curve, b is acceleration curve and a is the jerk curve.

Explanation of Solution

Note: Let f and $f^{\prime }$ be the graph of a function and its derivative on the same screen. Then $f^{\prime }\left(a\right)=0$ where a is x-coordinate of the point where f has horizontal tangent.

Graph:

The given graph is shown as in Figure 1,

Observation:

From Figure 1, the point where $a\left(t\right)=0$ is the same point where graph of $b\left(t\right)$ has horizontal slope.

Recall that the derivative of a function is zero where the function has a horizontal tangent.

The curve $a\left(t\right)$ is the derivative of the graph $b\left(t\right)$.

Thus, $b^{\prime }\left(t\right)=a\left(t\right)$. (1)

Observe the graph of b and c carefully.

The point where $b\left(t\right)=0$ is the same point where graph of $c\left(t\right)$ has horizontal slope.

Recall that the derivative of a function is zero where the function has a horizontal tangent.

The curve $b\left(t\right)$ is the derivative of the graph $c\left(t\right)$.

Thus, $c^{\prime }\left(t\right)=b\left(t\right)$. (2)

From (1) and (2), $b^{\prime }\left(t\right)=a\left(t\right)$ and $c^{\prime }\left(t\right)=b\left(t\right)$. This implies that, $b^{\prime }\left(t\right)=c^{″}\left(t\right)$.

There are two possibilities: $c^{\prime }\left(t\right)=d\left(t\right)$ or $d\left(t\right)=a^{\prime }\left(t\right)$

Observe the graph of d and a carefully.

The points where the graph of a has horizontal tangents, the functional value of d is not zero at that points.

This implies that $a^{\prime }\left(t\right)\ne d\left(t\right)$.

The only possibility is that a is the acceleration curve.

This implies $c^{\prime }\left(t\right)=d\left(t\right)$.

Thus d is position curve, c is velocity curve, b is acceleration curve and a is the jerk curve.