Chapter 7.1, Problem 47E

a.

To determine

To graph:

To graph the given initial value problem $\frac{dy}{dx}={\sec }^{2}x$ and $y=1$ when $x=\pi$

Answer to Problem 47E

  

Explanation of Solution

Given information:

The given initial value problem is $\frac{dy}{dx}={\sec }^{2}x$ and $y=1$ when $x=\pi$

Graph:

  

Now to find the general solution by finding the anti-derivative.

  $f\left(x\right)={\displaystyle \int {\sec }^{2}xdx=\tan x+c}$ [Where c is constant]

Now substituting the point $f\left(\pi \right)=1$ into the general solution and solve for c.

  $\begin{array}{l}f\left(\pi \right)=\tan \left(x\right)+c\\ 1=\tan \left(\pi \right)+c\\ 1=0+c\\ 1=c\end{array}$

So the solution of the initial value problem is

  $y=\tan x+1$

Interpretation:

We are only evaluating the part of the graph that passes through the point $\left(\pi ,1\right)$ , that is we draw only one period of the graph.

b.

To determine

To Show:

To show that the given graph is not the correct answer of part (a).

Answer to Problem 47E

The given graph is an incorrect graph to part (a).

Explanation of Solution

Given information:

The given graph is

  

Proof:

The student’s graph is an incorrect graph of the function because we are only evaluating a certain portion of the graph, the portion that passes through $\left(\pi ,1\right)$ .

The actual solution graph is only the portion of the graph that passes through the point and nothing more when the graph is discontinuous.

Hence, the given graph is not correct to part (a).