Chapter 2.1, Problem 4E

The point P(0.5, 0) lies on the curve y = cos πx.

(a) If Q is the point (x, cos πx), use your calculator to find the slope of the secant line PQ (.correct to six decimal places) for the following values of x:

(i) 0

(ii) 0.4

(iii) 0.49

(iv) 0.499

(v) 1

(vi) 0.6

(vii) 0.51

(viii) 0.501

(b) Using the result of part (a), guess the value of the slope of the tangent line to the curve at P(0.5, 0).

(c) Using the slope from part (b), find an equation of the tangent line to the curve at P(0.5, 0).

(d) Sketch the curve, two of the secant lines, and the tangent line.

To determine

To find: The slope of the secant line PQ for the following values of x.

Answer to Problem 4E

The slope of the secant line PQ for the following values of x is given below:

(i) The slope of the secant line PQ when $x=0$ is $-2$.

(ii) The slope of the secant line PQ when $x=0.4$ is $-3.090170$.

(iii) The slope of the secant line PQ when $x=0.49$ is $-3.141076$.

(iv) The slope of the secant line PQ when $x=0.499$ is $-3.141587$.

(v) The slope of the secant line PQ when $x=1$ is $-2$.

(vi) The slope of the secant line PQ when $x=0.6$ is $-3.090170$.

(vii) The slope of the secant line PQ when $x=0.51$ is $-3.141076$.

(viii) The slope of the secant line PQ when $x=0.501$ is $-3.141587$.

Explanation of Solution

Given:

The equation of the curve $y=\cos \pi x$.

The point P(0.5, 0) lies on the curve y.

The point Q is $\left(x,\cos \pi x\right)$.

Calculation:

The slope of the secant line between the points, P(0.5, 0) and Q$\left(x,\cos \pi x\right)$ is

$m_{PQ}=\frac{\cos \pi x-0}{x-0.5}$ (1)

Section (i):

Obtain the slope of the secant line PQ when $x=0$.

Substitute 0 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0\right)\\ =1\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0,1\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{1-0}{0-0.5}\\ =\frac{1}{-0.5}\\ =-2\end{array}$

Thus, the slope of the secant line PQ when $x=0$ is $-2$.

Section-(ii):

Obtain the slope of the secant line PQ when $x=0.4$.

Substitute 0.4 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.4\right)\\ =0.30901699\\ \approx 0.309017\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.4,0.309017\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{0.309017-0}{0.4-0.5}\\ =\frac{0.309017}{-0.1}\\ \approx -3.090170\end{array}$

Thus, the slope of the secant line PQ when $x=0.4$ is $-3.090170$.

Section-(iii):

Obtain the slope of the secant line PQ when $x=0.49$.

Substitute 0.49 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.49\right)\\ =0.0314107\\ \approx 0.031411\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.49,0.031411\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{0.031411-0}{0.49-0.5}\\ =\frac{0.031411}{-0.01}\\ \approx -3.141076\end{array}$

Thus, the slope of the secant line PQ when $x=0.49$ is $-3.141076$.

Section-(iv):

Obtain the slope of the secant line PQ when $x=0.499$.

Substitute 0.499 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.499\right)\\ =0.003141587\\ \approx 0.00314159\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.499,0.00314159\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{0.00314159-0}{0.499-0.5}\\ =\frac{0.00314159}{-0.001}\\ \approx -3.141587\end{array}$

Thus, the slope of the secant line PQ when $x=0.499$ is $-3.141587$.

Section-(v):

Obtain the slope of the secant line PQ when $x=1$.

Substitute 1 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(1\right)\\ =\cos \pi \\ =-1\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(1,-1\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{-1-0}{1-0.5}\\ =\frac{-1}{0.5}\\ \approx -2\end{array}$

Thus, the slope of the secant line PQ when $x=1$ is $-2$.

Section-(vi):

Obtain the slope of the secant line PQ when $x=0.6$.

Substitute 0.6 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.6\right)\\ =-0.30901699\\ \approx -0.309017\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.6,-0.309017\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{-0.309017-0}{0.6-0.5}\\ =\frac{-0.309017}{0.1}\\ \approx -3.090170\end{array}$

Thus, the slope of the secant line PQ when $x=0.6$ is $-3.090170$.

Section-(vii):

Obtain the slope of the secant line PQ when $x=0.51$.

Substitute 0.49 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.51\right)\\ =-0.0314107\\ \approx -0.031411\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.51,-0.031411\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{-0.031411-0}{0.51-0.5}\\ =\frac{-0.031411}{0.01}\\ \approx -3.141076\end{array}$

Thus, the slope of the secant line PQ when $x=0.51$ is $-3.141076$.

Section-(viii):

Obtain the slope of the secant line PQ when $x=0.501$.

Substitute 0.501 for x in $\cos \pi x$.

$\begin{array}{c}\cos \pi x=\cos \pi \left(0.501\right)\\ =-0.003141587\\ \approx -0.00314159\end{array}$

Substitute Q$\left(x,\cos \pi x\right)=\left(0.501,-0.00314159\right)$ in equation (1).

$\begin{array}{c}{m}_{PQ}=\frac{\cos \pi x-0}{x-0.5}\\ =\frac{-0.00314159-0}{0.501-0.5}\\ =\frac{-0.00314159}{0.001}\\ \approx -3.141587\end{array}$

Thus, the slope of the secant line PQ when $x=0.501$ is $-3.141587$.

To determine

To guess: The slope of the tangent line to the curve at P (0.5, 0).

Answer to Problem 4E

The estimated slope of the tangent line to the curve at P (0.5, 0) is $-\pi$.

Explanation of Solution

Formula used:

The slope of the tangent line is the limit of the slope of the secant line.

That is, $\underset{Q\to P}{\lim }\left(m_{PQ}\right)=m$ (2)

Calculation:

From part (a), the slope of the secant line for many values of x is closer to 1. Thus, the slope $m_{PQ}$ approaches to $-3.142$ or $-\pi$ .

Substitute $m_{PQ}=\frac{\cos \pi x-0}{x-0.5}$ in equation (2).

$\begin{array}{c}m=\underset{Q\to P}{\lim }\left(m_{PQ}\right)\\ =\underset{x\to 0.5}{\lim }\left(\frac{\cos \pi x-0}{x-0.5}\right)\\ =\left(\frac{\cos \pi \left(0.5\right)}{0.5-0.5}\right)\\ =\frac{0}{0}\end{array}$

Since $\frac{0}{0}$ is in indeterminate form, apply L’Hospital’s rule and obtain the limit.

$\begin{array}{c}\underset{x\to 0.5}{\lim }\left(\frac{\cos \pi x-0}{x-0.5}\right)=\underset{x\to 0.5}{\lim }\left(-\pi \sin \left(\pi x\right)\right)\\ =\left(-\pi \sin \left(\pi \left(0.5\right)\right)\right)\\ =-\pi \left(1\right)\\ =-\pi \end{array}$

Thus, the estimated slope of the tangent line to the curve at P (0.5, 0) is $-\pi$.

To determine

To find: The equation of the tangent line to the curve at P(0.5, 0).

Answer to Problem 4E

The equation of the tangent line to the curve at P(0.5, 0) is $\underset{\_}{y=-\pi x+\frac{1}{2}\pi }$

Explanation of Solution

Formula used:

The equation of the tangent line to the curve $y=f\left(x\right)$ at the point $\left(x_1,y_1\right)$:

$y-y_1=m\left(x-x_1\right)$ (3)

Calculation:

Substitute $m=-\pi$ and $\left(x_1,y_1\right)=\left(0.5,0\right)$ in equation (3).

$\begin{array}{c}y-0=-\pi \left(x-0.5\right)\\ y=-\pi x+0.5\pi \\ =-\pi x+\frac{1}{2}\pi \end{array}$

Thus, the equation of the tangent line to the curve at P(0.5,0) is $\underset{\_}{y=-\pi x+\frac{1}{2}\pi }$.

To determine

To sketch: The curve, two of the secant line, and the tangent line.

Explanation of Solution

Formula used:

Slope-intercept formula: $y-y_1=m_{PQ}\left(x-x_1\right)$

Calculation:

Obtain the secant line at $x=0$.

The slope of the secant line PQ when $x=0$ is $-2$.

Use the slope-point intercept formula to find a equation of the secant line.

Substitute $\left(x_1,y_1\right)=\left(0,1\right)$ and $m_{PQ}=-2$ in the Slope-point intercept formula.

$\begin{array}{c}y-1=-2\left(x-0\right)\\ y-1=-2x\\ y=-2x+1\end{array}$

Thus, the equation of the secant line at $x=0$ is $y=-2x+1$.

Obtain the secant line at $x=1$.

The slope of the secant line PQ when $x=1$ is $-2$.

Use the slope-point intercept formula to find a equation of the secant line.

Substitute $\left(x_1,y_1\right)=\left(1,-1\right)$ and $m_{PQ}=-2$ in the Slope-point intercept formula.

$\begin{array}{c}y-\left(-1\right)=-2\left(x-1\right)\\ y+1=-2x+2\\ y=-2x+2-1\\ =-2x+1\end{array}$

Thus, the equation of the secant line at $x=1$ is $y=-2x+1$.

Note that the equation of the secant line at $x=0$ and $x=1$ are the same.

Draw the curve $y=\cos \pi x$ , two of the secant lines $y=-2x+1$, and the tangent line $y=-\pi x+\frac{\pi }{2}$ as shown below in Figure 1.

Thus, the required sketch is obtained.