EXAM 2 Calculus 2

Question 1: Series Convergence and Divergence

a) The Ratio Test

How the Ratio Test Works

We compute the limit $L = \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right|$.

• If $L < 1$, the series **converges** absolutely.
• If $L > 1$, the series **diverges**.
• If $L = 1$, the test is inconclusive, and we must try another method.

Step-by-Step Solution

1. Identify $a_n$ and $a_{n+1}$: $$ a_n = \frac{n+1}{1.01^n} \quad \implies \quad a_{n+1} = \frac{(n+1)+1}{1.01^{n+1}} = \frac{n+2}{1.01^{n+1}} $$
2. Set up the ratio $\frac{a_{n+1}}{a_n}$: $$ \frac{a_{n+1}}{a_n} = \frac{\frac{n+2}{1.01^{n+1}}}{\frac{n+1}{1.01^n}} = \frac{n+2}{1.01^{n+1}} \cdot \frac{1.01^n}{n+1} $$
3. Simplify the expression: The exponential terms simplify: $\frac{1.01^n}{1.01^{n+1}} = \frac{1}{1.01}$. $$ \frac{a_{n+1}}{a_n} = \frac{n+2}{n+1} \cdot \frac{1}{1.01} $$
4. Compute the limit: $$ L = \lim_{n \to \infty} \left| \frac{n+2}{n+1} \cdot \frac{1}{1.01} \right| = \frac{1}{1.01} \cdot \lim_{n \to \infty} \frac{n+2}{n+1} $$ To solve the remaining limit, we can divide the numerator and denominator by the highest power of $n$: $$ \lim_{n \to \infty} \frac{1 + \frac{2}{n}}{1 + \frac{1}{n}} = \frac{1+0}{1+0} = 1 $$ So, the final limit is $L = \frac{1}{1.01} \cdot 1 = \frac{1}{1.01}$.
5. Conclusion: Since $L = \frac{1}{1.01} < 1$, the series $\sum_{n=1}^\infty \frac{n+1}{1.01^n}$ **converges** by the Ratio Test.

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b) The Limit Comparison Test

How the Limit Comparison Test Works

We choose a simpler series $\sum b_n$ whose convergence we know. We then compute the limit $L = \lim_{n \to \infty} \frac{a_n}{b_n}$.

• If $L$ is a finite, positive number ($0 < L < \infty$), then both series share the same fate: they either both converge or both diverge.

Step-by-Step Solution

1. Identify $a_n$ and choose a comparison series $b_n$: $$ a_n = \frac{\sqrt{n}}{n+1} $$ As discussed, we choose $b_n$ based on the dominant terms: $$ b_n = \frac{\sqrt{n}}{n} = \frac{1}{\sqrt{n}} = \frac{1}{n^{1/2}} $$
2. Analyze the comparison series $\sum b_n$:

   The series $\sum_{n=1}^\infty \frac{1}{n^{1/2}}$ is a **p-series** with $p = 1/2$. Since $p \le 1$, the series $\sum b_n$ **diverges**.

3. Compute the limit of the ratio: $$ L = \lim_{n \to \infty} \frac{a_n}{b_n} = \lim_{n \to \infty} \frac{\frac{\sqrt{n}}{n+1}}{\frac{1}{\sqrt{n}}} = \lim_{n \to \infty} \frac{\sqrt{n} \cdot \sqrt{n}}{n+1} = \lim_{n \to \infty} \frac{n}{n+1} $$ This limit is equal to 1.
4. Conclusion: Since $L=1$ (which is finite and positive) and we know that $\sum b_n$ diverges, the original series $\sum_{n=1}^\infty \frac{\sqrt{n}}{n+1}$ must also **diverge** by the Limit Comparison Test.

Question 2: Area Between Curves

Step 1: Find the Intersection Points (The Boundaries)

To find where the region begins and ends, we set the two functions equal to each other and solve for $x$.

$$ f(x) = g(x) \implies x^2 - 4x = 6 - x^2 $$ $$ 2x^2 - 4x - 6 = 0 $$ $$ x^2 - 2x - 3 = 0 $$ $$ (x-3)(x+1) = 0 $$

The intersection points are at $x = -1$ and $x = 3$. These will be our limits of integration, $a=-1$ and $b=3$.

Step 2: Determine the Top and Bottom Functions

In the interval $[-1, 3]$, we need to know which function has a greater y-value. We can test a point inside the interval, like $x=0$:

• $f(0) = 0^2 - 4(0) = 0$
• $g(0) = 6 - 0^2 = 6$

Since $g(0) > f(0)$, the function $g(x) = 6 - x^2$ is the **top function** and $f(x) = x^2 - 4x$ is the **bottom function** on this interval.

Step 3: Set Up and Evaluate the Integral

The area $A$ is given by the integral of (top - bottom):

$$ A = \int_{-1}^{3} \left[ (6 - x^2) - (x^2 - 4x) \right] dx $$ $$ A = \int_{-1}^{3} (6 + 4x - 2x^2) dx $$

Now, we find the antiderivative and apply the Fundamental Theorem of Calculus:

$$ A = \left[ 6x + 4\frac{x^2}{2} - 2\frac{x^3}{3} \right]_{-1}^{3} = \left[ 6x + 2x^2 - \frac{2}{3}x^3 \right]_{-1}^{3} $$

Evaluate at the upper and lower bounds:

$$ \text{At } x=3: \quad (6(3) + 2(3)^2 - \frac{2}{3}(3)^3) = (18 + 18 - 18) = 18 $$ $$ \text{At } x=-1: \quad (6(-1) + 2(-1)^2 - \frac{2}{3}(-1)^3) = (-6 + 2 + \frac{2}{3}) = -4 + \frac{2}{3} = -\frac{10}{3} $$

Finally, subtract the lower bound evaluation from the upper bound evaluation:

$$ A = 18 - \left(-\frac{10}{3}\right) = 18 + \frac{10}{3} = \frac{54}{3} + \frac{10}{3} = \frac{64}{3} $$

Interactive Visualization

The plot below shows the two functions and the shaded region whose area we calculated. You can see the intersection points and confirm that $g(x)$ is above $f(x)$ in the interval.

Question 4: Multivariable Extrema

Step 1: Find the Critical Points

First, we expand the function $f(x, y) = (2x + y)^2 + 2(x^2 - 1)^2$ for easier differentiation:

$$ f(x,y) = (4x^2 + 4xy + y^2) + 2(x^4 - 2x^2 + 1) = 4x^2 + 4xy + y^2 + 2x^4 - 4x^2 + 2 $$ $$ f(x,y) = 2x^4 + 4xy + y^2 + 2 $$

Now, compute the partial derivatives and set them to zero:

$$ \frac{\partial f}{\partial x} = 8x^3 + 4y = 0 \quad \implies \quad y = -2x^3 \quad \text{(1)} $$ $$ \frac{\partial f}{\partial y} = 4x + 2y = 0 \quad \implies \quad y = -2x \quad \text{(2)} $$

We have two expressions for $y$, so we can set them equal:

$$ -2x^3 = -2x \implies x^3 = x \implies x^3 - x = 0 \implies x(x^2-1) = 0 $$ This gives us three possible values for $x$: $x=0$, $x=1$, and $x=-1$. We use equation (2) to find the corresponding $y$ value for each:

• If $x=0$, then $y = -2(0) = 0$. Point: $(0,0)$.
• If $x=1$, then $y = -2(1) = -2$. Point: $(1,-2)$.
• If $x=-1$, then $y = -2(-1) = 2$. Point: $(-1,2)$.

Step 2: The Second Derivative Test

We calculate the second partial derivatives to find the discriminant $D = f_{xx}f_{yy} - (f_{xy})^2$.

$$ f_{xx} = \frac{\partial}{\partial x}(8x^3+4y) = 24x^2 $$ $$ f_{yy} = \frac{\partial}{\partial y}(4x+2y) = 2 $$ $$ f_{xy} = \frac{\partial}{\partial x}(4x+2y) = 4 $$ So, $D = (24x^2)(2) - (4)^2 = 48x^2 - 16$.

Now we test each critical point:

• At (0, 0): $D(0,0) = 48(0)^2 - 16 = -16$. Since $D < 0$, this is a **saddle point**.
• At (1, -2): $D(1,-2) = 48(1)^2 - 16 = 32$. Since $D > 0$, we check $f_{xx}$. $f_{xx}(1,-2) = 24(1)^2 = 24 > 0$. This is a **local minimum**.
• At (-1, 2): $D(-1,2) = 48(-1)^2 - 16 = 32$. Since $D > 0$, we check $f_{xx}$. $f_{xx}(-1,2) = 24(-1)^2 = 24 > 0$. This is also a **local minimum**.

Step 3: Check for Global Extrema

Here we can use a powerful observation. Look at the original form of the function:

$$ f(x, y) = (2x + y)^2 + 2(x^2 - 1)^2 $$

This is a sum of squared terms. Since any real number squared is non-negative, the smallest possible value for $f(x,y)$ is 0. Can our function actually equal 0? Let's check our local minima:

$$ f(1,-2) = (2(1) + (-2))^2 + 2((1)^2 - 1)^2 = (0)^2 + 2(0)^2 = 0 $$ $$ f(-1,2) = (2(-1) + 2)^2 + 2((-1)^2 - 1)^2 = (0)^2 + 2(0)^2 = 0 $$

Since the function value at these points is 0, and the function can never be less than 0, these two local minima are also **global minima**.

Interactive 3D Visualization

The plot below shows the surface of $f(x,y)$. You can clearly see the two "valleys" where the global minima occur and the saddle point between them at the origin.