Chapter 2.4, Problem 22E

(a)

To determine

To find: The slope of the curve $y=9-x^2$ at $x=a$.

Answer to Problem 22E

The slope of the curve $y=9-x^2$ at $x=a$ is $-2a$.

Explanation of Solution

Given information:

The curve is $y=9-x^2$ and point is $x=a$.

Calculation:

The formula for the slope of the curve $f\left(x\right)$ at $\left(a,f\left(a\right)\right)$ is $\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}$.

Substitute $a+h$ for $x$ in the function.

  $\begin{array}{c}f\left(x\right)=9-x^2\\ f\left(a+h\right)=9-{\left(a+h\right)}^2\\ f\left(a+h\right)=9-\left(a^2+h^2+2ah\right)\\ f\left(a+h\right)=9-a^2-h^2-2ah\end{array}$

Substitute $a$ for $x$ in the function.

  $\begin{array}{l}f\left(x\right)=9-x^2\\ f\left(a\right)=9-a^2\end{array}$

Substitute $9-a^2-h^2-2ah$ for $f\left(a+h\right)$ and $9-a^2$ for $f\left(a\right)$ in the formula for slope.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=\underset{h\to 0}{\lim }\frac{9-a^2-h^2-2ah-\left(9-a^2\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{\left(-h^2-2ah\right)}{h}\\ =\underset{h\to 0}{\lim }\frac{-h\left(h+2a\right)}{h}\\ =\underset{h\to 0}{\lim }\left(-\left(h+2a\right)\right)\end{array}$

Further simplify.

  $\begin{array}{c}\underset{h\to 0}{\lim }\frac{f\left(a+h\right)-f\left(a\right)}{h}=-\left(0+2a\right)\\ =-2a\end{array}$

Therefore, the slope of the curve $y=9-x^2$ at $x=a$ is $-2a$.

(b)

To determine

To find: The change in the tangent at $x=a$ as the value of $a$ changes.

Answer to Problem 22E

The slope of the tangent at the point $x=a$ decreases as the value $a$ increases.

Explanation of Solution

Given information:

The curve is $y=9-x^2$ and point is $x=a$.

Calculation:

As calculated in part (a), the slope of the curve $y=9-x^2$ at $x=a$ is equal to $-2a$.

The slope of the tangent at the point $x=a$ is equal to negative of twice of $a$.

Therefore, the slope of the tangent at the point $x=a$ decreases as the value $a$ increases.