Question 1

A window is of the form of a rectangle surmounted by a semicircle. The rectangle is made of clear glass, and the semicircle is made of tinted glass that transmits only half as much light per unit area as the clear glass does. The total perimeter is fixed at 400 cm. Find the width of the window that will admit most light. Neglect the thickness of the frame. Express the answer to the nearest cm.

Solution

Detailed Derivation

1. Variables and Constraints

Let:

• $w$ be the width of the window (and diameter of the semicircle).
• $h$ be the height of the rectangular part.
• $r$ be the radius of the semicircle, so $r=\frac{w}{2}$.

The total perimeter ($P$) is fixed at $400$ cm. The perimeter is the sum of the base ($w$), the two vertical sides ($2h$), and the semicircular arc ($\pi r$):

2. Light Admitted (Objective Function)

The clear glass is the rectangular area: $A_{\text{rect}}=wh$.

The tinted glass is the semicircular area: $A_{\text{semi}}=\frac{1}{2}\pi r^2=\frac{1}{2}\pi {\left(\frac{w}{2}\right)}^2=\frac{\pi w^2}{8}$.

Let $L_{\text{clear}}$ be the light transmitted per unit area by the clear glass. The tinted glass transmits $\frac{1}{2}{L}_{\text{clear}}$ per unit area.

The total light admitted ($L$) is proportional to the effective area ($A$):

3. Effective Area in Terms of Width ($w$)

First, we solve the perimeter constraint $(1)$ for $h$:

Next, substitute this expression for $h$ into the effective area equation $(2)$:

Since $\frac{\pi }{4}-\frac{\pi }{16}=\frac{4\pi -\pi }{16}=\frac{3\pi }{16}$, we have:

4. Maximization

To find the maximum effective area, we take the derivative of $A(w)$ with respect to $w$ and set it to zero ($\frac{dA}{dw}=0$):

Setting the derivative to zero:

5. Calculation

Using $\pi \approx 3.14159$:

Rounding to the nearest cm, the optimal width is:

Question 2

Let $f$ be a differentiable function. Use the definiton of derivative to show that

To show that $\frac{d}{dx}f(x^2)=2x{f}^{\prime }(x^2)$ using the definition of the derivative, we must apply the limit definition to the function $g(x)=f(x^2)$.

The definition of the derivative of a function $g(x)$ is:

In our case, $g(x)=f(x^2)$, so we need to find:

Solution

Derivation

1. Set up the Limit

Start by expanding the term $(x+h{)}^2$ inside the function $f$:

2. Introduce the Derivative Definition of $f$

We know that $f$ is a differentiable function, which means the derivative $f^{\prime }(a)$ exists and is defined as:

To use this, we let $a=x^2$ and the change in the input be $k=2xh+h^2$. As $h\to 0$, $k\to 0$ as well.

3. Algebraic Manipulation (Multiply and Divide by $k$)

We multiply and divide the expression inside the limit by $k=(2xh+h^2)$:

4. Separate the Limit Terms

We can split the limit of the product into the product of two limits:

5. Evaluate Limit 1

Let $k=2xh+h^2$. As $h\to 0$, $k\to 0$.

By the definition of the derivative of $f$, this limit is simply $f^{\prime }(x^2)$:

6. Evaluate Limit 2

Factor $h$ out of the numerator and simplify:

As $h\to 0$, the limit becomes:

7. Combine the Results

Substitute the results of Limit 1 and Limit 2 back into the main expression:

This completes the proof using the definition of the derivative. (Note: This is equivalent to applying the Chain Rule, where $\frac{d}{dx}f(g(x))=f^{\prime }(g(x))\cdot g^{\prime }(x)$ with $g(x)=x^2$.)

Question 3

Let $f(x)=x^4-3x+1=0$. Use intermediate value theorem there exists $c\in (1,2)$ so that $f(c)=0$. Use Newton's method with $x_0=1.5$ to find $c$, correct to 3 decimal places.

Solution

Verification using the Intermediate Value Theorem (IVT)

The Intermediate Value Theorem (IVT) states that if a function $f(x)$ is continuous on a closed interval $[a,b]$ and $k$ is any number between $f(a)$ and $f(b)$, then there exists at least one number $c$ in the open interval $(a,b)$ such that $f(c)=k$.

For the given polynomial function $f(x)=x^4-3x+1$, we want to show there exists a root $c\in (1,2)$ such that $f(c)=0$.

1. Continuity: Since $f(x)$ is a polynomial, it is continuous everywhere, including the interval $[1,2]$.

2. Evaluate Endpoints:

   • $f(1)=(1{)}^4-3(1)+1=1-3+1=-1$
   • $f(2)=(2{)}^4-3(2)+1=16-6+1=11$

3. Conclusion:

   Since $f(1)=-1$ and $f(2)=11$, and $0$ is between $-1$ and $11$ (i.e., $f(1)<0<f(2)$), by the IVT, there must exist at least one value $c$ in the interval $(1,2)$ such that $f(c)=0$.

Finding the Root using Newton's Method

Newton's Method (or the Newton-Raphson method) is an iterative process to approximate the roots of a real-valued function.

The iteration formula is:

1. Function and its Derivative

• Given function: $f(x)=x^4-3x+1$
• Derivative: $f^{\prime }(x)=4x^3-3$

2. Iterations

We start with the initial guess $x_0=1.5$. We stop when the successive approximations agree to 3 decimal places.

Since $x_3\approx 1.305904$ and $x_4\approx 1.305903$, they agree to at least 4 decimal places.

3. Final Answer

The root $c$ of $f(x)=0$, correct to 3 decimal places, is:

Question 4

Express the following as a definite integral ${\int }_0^{1}f(x)dx$ and find its exact value. Express your answer in the form $a\ln b$ where $a$, $b$ are to be determined.

(a) Let $n$ be a positive integer. Prove the reduction formula

Hence, evaluate ${\int }_0^{\pi /4}{\tan }^{4}xdx$. Express the answer in terms of $\pi$.

(b) Let $R$ be the region bounded by the curve $y=\frac{x}{1+x^3}$, $x=1$, and $y=0$. Find the volume when $R$ is rotated $2\pi$ radians about the $y$-axis. Express your answer in terms of $\pi$

Solution

1. Definite Integral and Exact Value

The given limit of a sum is a Riemann sum which can be expressed as a definite integral ${\int }_a^{b}f(x)dx$.

The standard form for a Riemann sum that converges to ${\int }_0^{1}f(x)dx$ is:

We have the expression:

1.1. Expressing as a Definite Integral

We need to manipulate the given sum into the standard form by factoring out terms involving $n$.

By comparing this with $\underset{n\to \mathrm{\infty }}{lim}\sum _{k=1}^{n}f\left(\frac{k}{n}\right)\frac{1}{n}$, we can make the following identification:

• $\frac{k}{n}\to x$
• $\frac{1}{n}\to dx$
• The integration limits are $a=0$ and $b=1$.

Therefore, the definite integral is:

where $f(x)=\frac{6x^2}{x^3+1}$.

1.2. Finding the Exact Value

We use u-substitution to evaluate the integral:

Let $u=x^3+1$.

Then the differential is $du=3x^{2}dx$.

We can rewrite the integral in terms of $du$:

Now, change the limits of integration:

• When $x=0$, $u=0^3+1=1$.
• When $x=1$, $u=1^3+1=2$.

The definite integral becomes:

Since $\ln 1=0$:

The exact value is $2\ln 2$.

The answer is in the form $a\ln b$, where $a=2$ and $b=2$.

(a) Trigonometric Reduction Formula

2.1. Prove the Reduction Formula

We want to prove:

We start by rewriting ${\tan }^{n}x$ using the identity ${\tan }^{2}x={\sec }^{2}x-1$:

The second integral is simply $I_{n-2}$. We focus on the first integral, $J$:

Use u-substitution:

Let $u=\tan x$.

Then $du={\sec }^{2}xdx$.

Assuming $n-2\ne -1$, or $n\ne 1$, we can integrate:

Substitute $J$ back into the expression for $I_n$:

This proves the reduction formula.

2.2. Evaluate ${\int }_0^{\pi /4}{\tan }^{4}xdx$

We use the reduction formula for $n=4$:

Now, we apply the formula again for $n=2$:

Substitute $I_2$ back into $I_4$:

Finally, we evaluate the definite integral from $0$ to $\frac{\pi }{4}$:

Since $\tan (\frac{\pi }{4})=1$ and $\tan (0)=0$:

The value is $\frac{\pi }{4}-\frac{2}{3}$.

(b) Volume of Revolution about the $y$-axis

We are asked to find the volume of the region $R$ bounded by $y=\frac{x}{1+x^3}$, $x=1$, and $y=0$ rotated $2\pi$ radians about the $y$-axis.

We will use the Cylindrical Shell Method.

3.1. Set up the Integral

Since the rotation is about the $y$-axis (a vertical axis), we integrate with respect to $x$.

The formula for the volume $V$ is:

• Integration Limits ($a,b$): The region is bounded by $x=0$ (from $y=0$ being the $x$-axis) and $x=1$. So, $a=0$ and $b=1$.
• Radius ($r$): The distance from the $y$-axis to a differential element is $r=x$.
• Height ($h$): The height of the shell is the function value: $h=y=\frac{x}{1+x^3}$.

3.2. Evaluation

We use u-substitution to evaluate the integral:

Let $u=1+x^3$.

Then the differential is $du=3x^{2}dx$, so $x^{2}dx=\frac{1}{3}du$.

Change the limits of integration:

• When $x=0$, $u=1+0^3=1$.
• When $x=1$, $u=1+1^3=2$.

The volume integral becomes:

Since $\ln 1=0$:

The volume is $\frac{2\pi }{3}\ln 2$.