PAGE 1

Optimization: Inscribed Rectangle in a Semicircle

A rectangle is inscribed in the upper half of the circle \(x^2 + y^2 = a^2\) as shown at right. Calculate the area of the largest such rectangle.

A. \(\frac{a^2}{2}\)
B. \(3a\sqrt{2}\)
C. \(2a^2\)
D. \(4a^2\)
E. \(a^2\) (Selected)

Figure: Semicircle with inscribed rectangle on axes.

Problem Setup

Coordinate graph of semicircle x^2+y^2=a^2 with rectangle of width 2x and height y. — Figure: Coordinate graph of semicircle \(x^2+y^2=a^2\) with rectangle of width \(2x\) and height \(y\).

height: \(y\)
length: \(2x\)

area: \(A = 2xy\)

need to eliminate one variable

Expressing Area as a Function of \(x\)

Circle equation: \(x^2 + y^2 = a^2\)

\[y^2 = a^2 - x^2\]\[y = \sqrt{a^2 - x^2} = (a^2 - x^2)^{1/2}\]

Substituting \(y\) into the area formula:

\[A(x) = 2x(a^2 - x^2)^{1/2}\]

Interval: \(0 \le x \le a\)

PAGE 2

Finding Critical Numbers

Find critical numbers, then compare \(A(x)\) at critical numbers and at end points.

\[A'(x) = (2x)(\frac{1}{2})(a^2 - x^2)^{-1/2}(-2x) + (a^2 - x^2)^{1/2}(2)\]

Chain Rule

\[= (-2x^2)(a^2 - x^2)^{-1/2} + 2(a^2 - x^2)^{1/2}\]\[= \frac{-2x^2}{(a^2 - x^2)^{1/2}} + 2(a^2 - x^2)^{1/2}\]

Solving for \(A'(x) = 0\)

\[A' = 0 \rightarrow \frac{-2x^2}{(a^2 - x^2)^{1/2}} + 2(a^2 - x^2)^{1/2} = 0\]\[2(a^2 - x^2)^{1/2} = \frac{2x^2}{(a^2 - x^2)^{1/2}}\]\[a^2 - x^2 = x^2\]\[2x^2 = a^2\]\[x^2 = \frac{1}{2}a^2\]\[x = \frac{1}{\sqrt{2}}a\]

\(x = \frac{1}{\sqrt{2}}a\)

\(-\frac{1}{\sqrt{2}}a\)

not inside interval of \(x\): \(0 \le x \le a\)

PAGE 3

\[ A(x) = 2x(a^2 - x^2)^{1/2} \]

Evaluating at specific points:

\[ A(0) = 0 \]

\[ A\left(\frac{1}{\sqrt{2}}a\right) = \frac{2}{\sqrt{2}}a \left(a^2 - \frac{1}{2}a^2\right)^{1/2} = \sqrt{2}a \left(\frac{1}{2}a^2\right)^{1/2} = \sqrt{2}a \cdot \frac{1}{\sqrt{2}}a = a^2 \]

\[ A(a) = 0 \]

PAGE 4

Mean Value Theorem Problem

Given that \( f(x) \) is differentiable for all \( x \), \( f(2) = 4 \), and \( f(7) = 10 \), then the Mean Value Theorem states that there is a number \( c \) such that:

A. \( 2 < c < 7 \) and \( f'(c) = \frac{6}{5} \)
B. \( 2 < c < 7 \) and \( f'(c) = \frac{5}{6} \)
C. \( 4 < c < 10 \) and \( f'(c) = \frac{6}{5} \)
D. \( 2 < c < 7 \) and \( f'(c) = 0 \)
E. \( 4 < c < 10 \) and \( f'(c) = 0 \)

Mean Value Theorem Definition

\( f(x) \) continuous on \( [a, b] \)

\( f(x) \) differentiable on \( (a, b) \)

then there is a "\( c \)" such that \( a < c < b \)

\[ f'(c) = \frac{f(b) - f(a)}{b - a} \]

Somewhere between \( a \) and \( b \) there is at least one place \( c \) where the tangent line slope = secant line slope through endpoints.

Figure: Coordinate graph showing a curve from a to b with a dashed secant line and a parallel tangent line at point c.

PAGE 5

Mean Value Theorem Application

Given the following information about a function:

Know:

\( f(2) = 4 \)
\( f(7) = 10 \)

For the interval \( 2 \le x \le 7 \) (where \( a = 2 \) and \( b = 7 \)):

Somewhere between \( x = 2 \) and \( x = 7 \) there is a place where:

\[ f' = \frac{f(7) - f(2)}{7 - 2} = \frac{10 - 4}{7 - 2} = \frac{6}{5} \]

PAGE 6

Function Analysis: \( g(x) = 4x^3 - 3x^4 \)

Which of the following is/are true about the function \( g(x) = 4x^3 - 3x^4 \)?

\( g \) is decreasing for \( x > 1 \).
\( g \) has a relative extreme value at \( (0,0) \).
The graph of \( g \) is concave up for all \( x < 0 \).

Options: A. (1), (2) and (3) | B. only (2) | C. only (1) | D. (1) and (2) | E. (1) and (3)

First Derivative Analysis

\[ g(x) = 4x^3 - 3x^4 \]\[ g'(x) = 12x^2 - 12x^3 = 0 \]\[ 12x^2(1 - x) = 0 \implies x = 0, x = 1 \]

Sign of \( g'(x) \):

Figure: Sign chart for g'(x) with critical points at 0 and 1, showing positive slope then negative.

\( x = 0 \) (no sign change), \( x = 1 \) (sign change)

Increasing: \( (-\infty, 0), (0, 1) \)
Decreasing: \( (1, \infty) \) \( \rightarrow \) Statement (1) is correct

(2) is FALSE because for an extreme value to exist at \( x = 0 \) there has to be a sign change in \( g' \) across \( x = 0 \).

Second Derivative Analysis

\[ g''(x) = 24x - 36x^2 = 0 \]\[ 12x(2 - 3x) = 0 \implies x = 0, x = 2/3 \]

Sign of \( g'' \):

Figure: Sign chart for g''(x) with points 0 and 2/3, showing concave down, then up, then down.

\( x = 0 \), \( x = 2/3 \)

(3) is FALSE (Graph is concave down for \( x < 0 \))

PAGE 7

Limit Evaluation using L'Hôpital's Rule

Problem Statement:

Evaluate the limit: \[ \lim_{x \to 0} \frac{2x - \sin^{-1} x}{2x + \tan^{-1} x} = \]

A. 1/2B. 2C. 1/3D. 1E. 0

Step-by-Step Solution

\[ \lim_{x \to 0} \frac{2x - \sin^{-1} x}{2x + \tan^{-1} x} \xrightarrow{x=0} \frac{0}{0} \]

L'Hôpital's Rule can be used (the other form is \(\frac{\infty}{\infty}\))

\[ \text{L'Hôpital's Rule: } \lim_{x \to a} \frac{f(x)}{g(x)} = \lim_{x \to a} \frac{f'(x)}{g'(x)} \text{ if limit } \to \frac{0}{0} \text{ or } \frac{\infty}{\infty} \]

Applying the rule by differentiating the numerator and denominator:

\[ \lim_{x \to 0} \frac{2 - \frac{1}{\sqrt{1-x^2}}}{2 + \frac{1}{1+x^2}} \xrightarrow{x=0} \frac{2-1}{2+1} = \frac{1}{3} \]

Do NOT use L'Hôpital's Rule if not \(\frac{0}{0}\) or \(\frac{\infty}{\infty}\)

PAGE 8

Differentiation of an Integral

Problem Statement:

Find \( \frac{d}{dx} \int_{1}^{2x} \sqrt{t^2 + 1} dt \) at \( x = \sqrt{2} \).

A. 6B. 3C. \(\sqrt{2}\)D. \(\sqrt{4x^2 + 1}\)E. \(\frac{1}{2\sqrt{3}}\)

Conceptual Framework

The derivative of an integral leads to the Fundamental Theorem of Calculus (Part 1).

FTC 1:

\[ \frac{d}{dx} \int_{a}^{x} f(t) dt = f(x) \]

Here, the upper limit is not just \(x\), so we need to use the Chain Rule.

Application

Let \( u = 2x \)

Then:

\[ \frac{d}{dx} \int_{1}^{2x} \sqrt{t^2 + 1} dt = \frac{d}{dx} \int_{1}^{u} \sqrt{t^2 + 1} dt \]

Applying the Chain Rule \( \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \):

\[ \left( \frac{d}{du} \int_{1}^{u} \sqrt{t^2 + 1} dt \right) \left( \frac{d}{dx} 2x \right) \]

FTC 1 applied to the first term

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Substitution Evaluation Continued

\[ = (\sqrt{u^2 + 1})(2) \quad \text{where } u = 2x \]

\[ = 2\sqrt{(2x)^2 + 1} = 2\sqrt{4x^2 + 1} \]

At \( x = \sqrt{2} \) we get:

\[ 2\sqrt{4(\sqrt{2})^2 + 1} = 6 \]

PAGE 10

Definite Integral with Substitution

\[ \int_{3}^{4} x\sqrt{25 - x^2} dx = \]

A. 0B. -37C. \(\frac{37}{3}\)D. \(-\frac{74}{3}\)E. \(\frac{7}{12}\)

\[ \int_{3}^{4} x\sqrt{25 - x^2} dx \]

need to use a substitution

\[ = \int_{3}^{4} x (25 - x^2)^{1/2} dx \]

Compare \( x \) and \( 25 - x^2 \).

The derivative of \( 25 - x^2 \) is \( -2x \) which is a constant multiple of \( x \), so we let:

\( u = 25 - x^2 \)

\( \frac{du}{dx} = -2x \)

\( du = -2x \, dx \)

now we adjust the integration limits

PAGE 11

Definite Integration with Substitution

Changing the Limits of Integration

When performing a substitution in a definite integral, it is often easier to change the limits of integration to match the new variable.

Old upper limit: \( x = 4 \rightarrow u = 25 - x^2 = 25 - 16 = 9 \)
Old lower limit: \( x = 3 \rightarrow u = 25 - x^2 = 25 - 9 = 16 \)

Substitution Process

Original integral with annotations:

\[ \int_{3}^{4} (25 - x^2)^{1/2} (x) dx \]

Let \( u = 25 - x^2 \), then \( du = -2x dx \), so \( x dx = -\frac{1}{2} du \).

Substituting the new limits and the variable \( u \):

\[ = \int_{16}^{9} -\frac{1}{2} u^{1/2} du = -\frac{1}{2} \int_{16}^{9} u^{1/2} du = -\frac{1}{2} \left( \frac{u^{3/2}}{3/2} \right) \bigg|_{16}^{9} \]

DO NOT go back to x

Evaluation

\[ = -\frac{1}{2} \left( \frac{2}{3} u^{3/2} \right) \bigg|_{16}^{9} = -\frac{1}{2} \left( \frac{2}{3} \cdot 9^{3/2} \right) - \left( -\frac{1}{2} \left( \frac{2}{3} \cdot 16^{3/2} \right) \right) \]\[ = -\frac{1}{2} \left( \frac{2}{3} \cdot 27 \right) + \frac{1}{2} \left( \frac{2}{3} \cdot 64 \right) \]\[ = -\frac{1}{3} \cdot 27 + \frac{1}{3} \cdot 64 = -9 + \frac{64}{3} = \frac{37}{3} \]

PAGE 12

Suppose that the mass of a radioactive substance decays from 18 gms to 2 gms in 2 days. How long will it take for 12 gms of this substance to decay to 4 gms?

A. \( \frac{\ln 3}{\ln 2} \) days

B. 1 day (Correct)

C. \( \frac{\ln 2}{\ln 3} \) days

D. 2 days

E. \( (\ln 3)^2 \) days

Exponential Growth/Decay Model

The general formula is: \( y(t) = y_0 e^{kt} \)

\( y_0 = 18 \)
\( t \): time in days
\( y(2) = 2 \)

Finding the Decay Constant (k)

\[ y(t) = 18 e^{kt} \]\[ y(2) = 18 e^{k \cdot 2} = 2 \rightarrow e^{2k} = \frac{1}{9} \rightarrow \ln e^{2k} = \ln \frac{1}{9} \]\[ 2k = \ln \frac{1}{9} \rightarrow k = \frac{1}{2} \ln \frac{1}{9} \]

Solving for the New Condition

Find \( t \) when the mass decays from 12 gms to 4 gms:

\( 12 \rightarrow 4 \)
\( y_0 = 12 \)
\( y(t) = 4 \)

\[ y(t) = y_0 e^{kt} \]\[ 4 = 12 e^{\left( \frac{1}{2} \ln \frac{1}{9} \right) t} \]

PAGE 13

Solving Exponential Equations

The following steps demonstrate how to solve for the variable \(t\) in an exponential equation involving natural logarithms.

\[ \frac{1}{3} = e^{\ln\left(\frac{1}{9}\right)^{1/2} t} \]

Note: A crossed-out intermediate step \(\frac{1}{3} = \left(\frac{1}{9}\right)^{1/2} t\) is present in the original notes, as well as a crossed-out identity \(e^{\ln x} = x\).

Taking the natural logarithm of both sides:

\[ \ln \frac{1}{3} = \ln e^{\ln\left(\frac{1}{9}\right)^{1/2} t} \]

Using the property \(\ln e^x = x\):

\[ \ln \frac{1}{3} = \ln\left(\frac{1}{9}\right)^{1/2} t \]

Simplifying the right side since \(\left(\frac{1}{9}\right)^{1/2} = \frac{1}{3}\):

\[ \ln\left(\frac{1}{3}\right) = \ln\left(\frac{1}{3}\right) t \]

so \( t = 1 \)