PAGE 1

Related Rates: Rectangle Area

The length of a rectangle is increasing at a constant rate of 8 cm/s and its width is decreasing at a constant rate of 3 cm/s. How fast, in cm²/s, is the area of the rectangle increasing at the moment when the length is 20 cm and the width is 10 cm?

do NOT plug in until taking derivative

A. 140
B. 130
C. 40
D. 24
E. 20

Figure: A simple rectangle diagram with the horizontal side labeled x and the vertical side labeled y.

x: length
y: width

given:
\[ \frac{dx}{dt} = 8 \]
\[ \frac{dy}{dt} = -3 \]

find:
\[ \frac{dA}{dt} \text{ when } x=20, y=10 \]

\[ A = xy \]

both $ x $ and $ y $ are functions of $ t $ (time)

\[ \frac{d}{dt} A = \frac{d}{dt} (xy) = x \frac{dy}{dt} + y \frac{dx}{dt} \]

now we can plug in numbers

\[ = x(-3) + y(8) \]

\[ = -3x + 8y \]

at the moment when $ x=20, y=10 $

\[ \frac{dA}{dt} = -60 + 80 = \boxed{20} \]

increasing at 20 cm²/s

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Linear Approximation

Use a linear approximation to estimate the value of $ e^{-0.01} $

A. 1.001
B. 1.01
C. 0.9
D. 0.99
E. 0.999

linear approx: \[ L = f(a) + f'(a)(x-a) \]

identify $ f(x) $: $ e^x $

a: a convenient number where we know $ f(a) $ easily and close to where we want to be

here, we use $ a=0 $ since we know $ e^0 = 1 $ and $ 0 $ is close to $ -0.01 $

\[ L = f(a) + f'(a)(x-a) \]

$ f(a) = e^0 = 1 $
$ f'(x) = e^x $
$ f'(a) = e^0 = 1 $

\[ L = 1 + (1)(x-0) = 1 + x \approx e^x \]

\[ e^{-0.01} \approx 1 + (-0.01) = 0.99 \]

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Absolute Extrema on a Closed Interval

Note: Always work with a given closed interval.

Find the absolute maximum and minimum values of the function $ f(x) = \cos x + \sin x $ on the interval $ [0, \pi] $.

A. max: $ \sqrt{2} $; min: $ -1 $

B. max: $ \frac{\pi}{4} $; min: $ \pi $

C. max: 2; min: -2

D. max: 1; min: -1

E. max: $ \pi $; min: 0

Procedure:

Find critical pts $ \rightarrow f' = 0 $ or $ f' \text{ DNE} $
Then compare $ f(x) $ at critical pts and at the end points of interval

We do NOT use first or second derivative test with absolute max/min.

Step-by-Step Solution

$ f'(x) = -\sin x + \cos x = 0 $

Solve $ \cos x = \sin x $ on $ [0, \pi] $

or $ \tan x = 1 $ on $ [0, \pi] $

$ x = \pi/4 $ only crit. pt.

Compare $ f(x) = \cos x + \sin x $ at $ x = 0, \pi/4, \pi $:

$ f(0) = 1 $

$ f(\pi/4) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.4 \quad \text{ (max)} $

$ f(\pi) = -1 \quad \text{ (min)} $

PAGE 4

Evaluating Limits with L'Hospital's Rule

$ \lim_{x \to \pi} \frac{\cos x + \sin 2x + 1}{x^2 - \pi^2} = \frac{-1 + 0 + 1}{\pi^2 - \pi^2} \rightarrow \frac{0}{0} $

A. $ \frac{1}{2\pi} $

B. $ \frac{1}{\pi} $

C. 1

D. $ -\frac{1}{\pi} $

E. $ -\frac{1}{2\pi} $

L'Hospital's Rule

When limit seems to go to $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $
Take the deriv. of top and bottom and then re-evaluate limit
Repeat until no longer $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $

Application

$ \stackrel{L}{=} \lim_{x \to \pi} \frac{-\sin x + 2\cos 2x}{2x} $

Now try $ x = \pi $:

$ = \frac{-\sin(\pi) + 2\cos(2\pi)}{2\pi} = \frac{2}{2\pi} = \frac{1}{\pi} $

Do NOT apply l'Hospital's Rule because not $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $.

PAGE 5

Evaluating Indeterminate Limits

\[ \lim_{x \to 0} (1 + 2x)^{\cot x} = \]

As $ x \to 0 $, the expression takes the form $ 1^{\infty} $, which is an indeterminate form.

Strategy: Turn into $ \frac{0}{0} $ or $ \frac{\infty}{\infty} $ then use L'Hôpital's Rule.

Let $ y = (1 + 2x)^{\cot x} $. We want to find $ \lim_{x \to 0} y $.

Taking the natural logarithm of both sides:

\[ \ln y = \ln (1 + 2x)^{\cot x} \]\[ \ln y = (\cot x) \ln (1 + 2x) \]\[ \ln y = \frac{\ln (1 + 2x)}{\tan x} \]

Multiple Choice Options:

A. 0
B. 1
C. 2
D. $ e $
E. $ e^2 $

Note: as $ x \to 0 $, $ \frac{\ln (1 + 2x)}{\tan x} \to \frac{0}{0} $. So we can use L'Hôpital's Rule to find $ \lim_{x \to 0} \ln y $.

PAGE 6

Applying L'Hôpital's Rule

\[ \lim_{x \to 0} \ln y = \lim_{x \to 0} \frac{\ln (1 + 2x)}{\tan x} \]\[ \stackrel{L}{=} \lim_{x \to 0} \frac{\frac{2}{1 + 2x}}{\sec^2 x} \]

Recall: $ \sec x = \frac{1}{\cos x} $ and $ \cos(0) = 1 $

If $ x = 0 $:

\[ = \frac{\frac{2}{1}}{1} = 2 \]

Final Result

Not done: $ \lim_{x \to 0} \ln y = 2 $ but we want $ \lim_{x \to 0} y $.

\[ y = e^{\ln y} = e^2 \]

The correct answer is E.

PAGE 7

Find the locations ( $x$ values) of the inflection points of $f(x) = e^{-x^2}$. Note that its first and second derivatives are: $f'(x) = -2xe^{-x^2}$, $f''(x) = (4x^2 - 2)e^{-x^2}$.

A. $x = 1$ and $x = -1$
B. $x = 2$ and $x = -2$
C. $x = 0$
D. $x = \sqrt{2}$ and $x = -\sqrt{2}$
E. $x = \frac{1}{\sqrt{2}}$ and $x = -\frac{1}{\sqrt{2}}$

inflection pts: $f'' = 0$ or $f''$ DNE

and $f''$ changes sign

\[f'' = (4x^2 - 2)e^{-x^2} = 0\]

\[4x^2 - 2 = 0 \quad \text{or} \quad e^{-x^2} = 0\]

$e^{-x^2} = 0$ never

\[x^2 = \frac{1}{2}\]

\[x = \frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\]

Sign Analysis

Sign chart for f'' with roots at -\frac{1}{\sqrt{2}} and \frac{1}{\sqrt{2}}, showing concavity changes. — Figure: Sign chart for $f''$ with roots at $-\frac{1}{\sqrt{2}}$ and $\frac{1}{\sqrt{2}}$, showing concavity changes.

Sign change across both, so both are where inflection pts are.

\[f''(-2) > 0 \quad f''(0) = -2 \quad f''(2) > 0\]

PAGE 8

A six-sided box is to have four clear plastic sides, a wooden square top, and a wooden square bottom. The volume of the box must be $24 \text{ ft}^3$. Plastic costs $\$1$ per $\text{ft}^2$ and wood costs $\$3$ per $\text{ft}^2$. Find the dimensions of the box which minimize cost.

A. $2 \text{ ft} \times 2 \text{ ft} \times 6 \text{ ft}$
B. $\sqrt{6} \text{ ft} \times \sqrt{6} \text{ ft} \times 4 \text{ ft}$
C. $\sqrt[3]{4} \text{ ft} \times \sqrt[3]{4} \text{ ft} \times 6\sqrt[3]{4} \text{ ft}$
D. $\sqrt[3]{3} \text{ ft} \times \sqrt[3]{3} \text{ ft} \times 8\sqrt[3]{3} \text{ ft}$
E. $2\sqrt[3]{2} \text{ ft} \times 2\sqrt[3]{2} \text{ ft} \times 3\sqrt[3]{2} \text{ ft}$

Figure: Diagram of a rectangular prism with square base sides labeled x and height labeled y.

Optimization Setup

$\text{volume} = x^2y = 24$ (constraint)

$\text{cost} = (x^2)(3) + (x^2)(3) + (4)(xy)(1)$

basetopfour sides

\[= 6x^2 + 4xy\]

Solving for Dimensions

Eliminate $y$ using constraint:

\[x^2y = 24 \rightarrow y = \frac{24}{x^2}\]

\[C(x) = 6x^2 + 4x\left(\frac{24}{x^2}\right) = 6x^2 + \frac{96}{x}\]

\[C'(x) = 12x - \frac{96}{x^2} = 0 \rightarrow 12x = \frac{96}{x^2} \rightarrow x^3 = \frac{96}{12} = 8 \rightarrow x = 2\]

\[y = \frac{24}{x^2} = 6\]

Verification

Check $x=2$ minimizes cost using 2nd deriv. test:

\[C'' = 12 + \frac{192}{x^3}\]

\[C''(2) > 0 \text{ so } x=2 \text{ min } C\]

PAGE 9

F17 E3 #4

\[f(x) = \frac{\ln x}{x^2} \text{ on } [\frac{1}{e}, e]\]

Find abs. max/min

Find critical numbers: $f' = 0$ or $f' \text{ DNE}$

\[f'(x) = \frac{(x^2)(\frac{1}{x}) - (\ln x)(2x)}{(x^2)^2} = \frac{x - 2x \ln x}{x^4}\]

\[f' = 0 \rightarrow x - 2x \ln x = 0\]\[x(1 - 2 \ln x) = 0\]

$x \neq 0$

discard because not in $[\frac{1}{e}, e]$

or $1 - 2 \ln x = 0$

$\ln x = \frac{1}{2}$

$x = e^{1/2}$ keep

\[f' \text{ DNE} \rightarrow x^4 = 0 \rightarrow x \neq 0\]

$f(\frac{1}{e}) = \frac{\ln \frac{1}{e}}{(\frac{1}{e})^2} = \frac{-1}{\frac{1}{e^2}} = -e^2$

min

$f(e^{1/2}) = \frac{\ln e^{1/2}}{(e^{1/2})^2} = \frac{1/2}{e} = \frac{1}{2e}$

max

$f(e) = \frac{\ln e}{e^2} = \frac{1}{e^2}$

PAGE 10

F17 E3 #4

\[\sqrt{9.1}\]

Identify $f(x)$:

$f(x) = \sqrt{x} = x^{1/2}$

Identify $a$:

close to 9.1 and $\sqrt{a}$ is known

so, $a = 9$

$L = f(a) + f'(a)(x - a)$

$f(a) = \sqrt{9} = 3$

$f'(x) = \frac{1}{2}x^{-1/2} = \frac{1}{2\sqrt{x}}$

$f'(a) = f'(9) = \frac{1}{2\sqrt{9}} = \frac{1}{6}$

\[L = 3 + \frac{1}{6}(x - 9) \approx \sqrt{x}\]

\[\sqrt{9.1} \approx 3 + \frac{1}{6}(0.1) = 3 + \frac{1}{60} = 3\frac{1}{60}\]