We are asked to solve the following problems: a. Find the equation of the tangent plane to $z = \ln(2x+y)$ at the point $(1,2)$. b. State that the series $\sum_{x=1}^\infty \frac{1}{x^p}$ is a p-series and converges if $p>1$ and diverges if $0 < p \le 1$. c. Given $u = (x^2+y^2+z^2)^{\frac{1}{2}}$, find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$. d. If $a_n > 0$ for all $n$, prove that if $\frac{a_{n+1}}{a_n} < k < 1$, where k is a constant, then $\sum a_n$ is convergent. Hence, show that $\sum_{n=1}^\infty \frac{2^n}{n^3+1}$ diverges. e. Use the 5-segment method to integrate $f(x) = 2+5x+3x^2+2x^3$ from $a=0$ to $b=2.5$ using: [i] Mid-ordinate rule. Determine the relative error. [ii] Simpson's one-third rule. Determine the relative error. f. Find the slopes of the traces to $z = 10-4x^2-y^2$ at the point $(1,2)$. g. Find the radius of convergence for the power series $\sum_{n=1}^\infty \frac{n(-1)^n}{4^n} (x+3)^n$.

1. Problem Description

We are asked to solve the following problems:

a. Find the equation of the tangent plane to $z = \ln(2x+y)$ at the point $(1,2)$.

b. State that the series $\sum_{x=1}^\infty \frac{1}{x^p}$ is a p-series and converges if $p>1$ and diverges if $0 < p \le 1$.

c. Given $u = (x^2+y^2+z^2)^{\frac{1}{2}}$, find $\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2}$.

d. If $a_n > 0$ for all $n$, prove that if $\frac{a_{n+1}}{a_n} < k < 1$, where k is a constant, then $\sum a_n$ is convergent. Hence, show that $\sum_{n=1}^\infty \frac{2^n}{n^3+1}$ diverges.

e. Use the 5-segment method to integrate $f(x) = 2+5x+3x^2+2x^3$ from $a=0$ to $b=2.5$ using:

[i] Mid-ordinate rule. Determine the relative error.

[ii] Simpson's one-third rule. Determine the relative error.

f. Find the slopes of the traces to $z = 10-4x^2-y^2$ at the point $(1,2)$.

g. Find the radius of convergence for the power series $\sum_{n=1}^\infty \frac{n(-1)^n}{4^n} (x+3)^n$.

2. Solution Steps

a. To find the equation of the tangent plane to $z = \ln(2x+y)$ at $(1,2)$, we need to find the partial derivatives with respect to $x$ and $y$.

$\frac{\partial z}{\partial x} = \frac{2}{2x+y}$ and $\frac{\partial z}{\partial y} = \frac{1}{2x+y}$.

At $(1,2)$, $z = \ln(2(1)+2) = \ln(4)$.

$\frac{\partial z}{\partial x}(1,2) = \frac{2}{2(1)+2} = \frac{2}{4} = \frac{1}{2}$.

$\frac{\partial z}{\partial y}(1,2) = \frac{1}{2(1)+2} = \frac{1}{4}$.

The equation of the tangent plane is given by

$z - z_0 = \frac{\partial z}{\partial x}(x-x_0) + \frac{\partial z}{\partial y}(y-y_0)$.

$z - \ln(4) = \frac{1}{2}(x-1) + \frac{1}{4}(y-2)$

$z = \frac{1}{2}x + \frac{1}{4}y - \frac{1}{2} - \frac{1}{2} + \ln(4)$

$z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. The statement is a known result about p-series.

c. $u = (x^2+y^2+z^2)^{\frac{1}{2}}$.

$\frac{\partial u}{\partial x} = \frac{1}{2}(x^2+y^2+z^2)^{-\frac{1}{2}} (2x) = \frac{x}{\sqrt{x^2+y^2+z^2}}$.

$\frac{\partial^2 u}{\partial x^2} = \frac{\sqrt{x^2+y^2+z^2} - x \frac{2x}{2\sqrt{x^2+y^2+z^2}}}{x^2+y^2+z^2} = \frac{x^2+y^2+z^2 - x^2}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{y^2+z^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}$.

Similarly, $\frac{\partial^2 u}{\partial y^2} = \frac{x^2+z^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}$ and $\frac{\partial^2 u}{\partial z^2} = \frac{x^2+y^2}{(x^2+y^2+z^2)^{\frac{3}{2}}}$.

$\frac{\partial^2 u}{\partial x^2} + \frac{\partial^2 u}{\partial y^2} + \frac{\partial^2 u}{\partial z^2} = \frac{y^2+z^2+x^2+z^2+x^2+y^2}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{2(x^2+y^2+z^2)}{(x^2+y^2+z^2)^{\frac{3}{2}}} = \frac{2}{\sqrt{x^2+y^2+z^2}}$.

d. This is the Ratio Test. For the series $\sum_{n=1}^\infty \frac{2^n}{n^3+1}$,

$\frac{a_{n+1}}{a_n} = \frac{2^{n+1}}{(n+1)^3+1} \cdot \frac{n^3+1}{2^n} = 2 \frac{n^3+1}{(n+1)^3+1} = 2 \frac{n^3+1}{n^3+3n^2+3n+2}$.

$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} = 2 \lim_{n\to\infty} \frac{1+\frac{1}{n^3}}{1+\frac{3}{n}+\frac{3}{n^2}+\frac{2}{n^3}} = 2 > 1$. Since the limit is greater than 1, the series diverges.

e. We have $f(x) = 2+5x+3x^2+2x^3$. $a=0$, $b=2.5$, and the number of segments is 5, so the width of each segment is $h = \frac{2.5-0}{5} = 0.5$.

[i] Mid-ordinate rule:

The midpoints are $0.25, 0.75, 1.25, 1.75, 2.25$.

$f(0.25) = 2 + 5(0.25) + 3(0.25)^2 + 2(0.25)^3 = 2 + 1.25 + 0.1875 + 0.03125 = 3.46875$

$f(0.75) = 2 + 5(0.75) + 3(0.75)^2 + 2(0.75)^3 = 2 + 3.75 + 1.6875 + 0.84375 = 8.28125$

$f(1.25) = 2 + 5(1.25) + 3(1.25)^2 + 2(1.25)^3 = 2 + 6.25 + 4.6875 + 3.90625 = 16.84375$

$f(1.75) = 2 + 5(1.75) + 3(1.75)^2 + 2(1.75)^3 = 2 + 8.75 + 9.1875 + 10.71875 = 30.65625$

$f(2.25) = 2 + 5(2.25) + 3(2.25)^2 + 2(2.25)^3 = 2 + 11.25 + 15.1875 + 22.78125 = 51.21875$

Integral $\approx h \sum_{i=1}^5 f(x_i) = 0.5(3.46875+8.28125+16.84375+30.65625+51.21875) = 0.5(110.46875) = 55.234375$

Exact integral: $\int_0^{2.5} (2+5x+3x^2+2x^3) dx = [2x+\frac{5x^2}{2} + x^3 + \frac{x^4}{2}]_0^{2.5} = 2(2.5) + \frac{5(2.5)^2}{2} + (2.5)^3 + \frac{(2.5)^4}{2} = 5 + 15.625 + 15.625 + 19.53125 = 55.78125$.

Relative error = $\frac{|55.78125 - 55.234375|}{55.78125} = \frac{0.546875}{55.78125} \approx 0.009803 = 0.9803\%$.

[ii] Simpson's one-third rule:

$h = 0.5$.

$x_0 = 0, x_1 = 0.5, x_2 = 1, x_3 = 1.5, x_4 = 2, x_5 = 2.5$.

$f(0) = 2, f(0.5) = 2 + 5(0.5) + 3(0.5)^2 + 2(0.5)^3 = 2 + 2.5 + 0.75 + 0.25 = 5.5$

$f(1) = 2 + 5(1) + 3(1)^2 + 2(1)^3 = 2 + 5 + 3 + 2 = 12$

$f(1.5) = 2 + 5(1.5) + 3(1.5)^2 + 2(1.5)^3 = 2 + 7.5 + 6.75 + 6.75 = 23.0$

$f(2) = 2 + 5(2) + 3(2)^2 + 2(2)^3 = 2 + 10 + 12 + 16 = 40$

$f(2.5) = 2 + 5(2.5) + 3(2.5)^2 + 2(2.5)^3 = 2 + 12.5 + 18.75 + 31.25 = 64.5$

$\int_0^{2.5} f(x) dx \approx \frac{h}{3}[f(x_0) + 4f(x_1) + 2f(x_2) + 4f(x_3) + 2f(x_4) + f(x_5)] = \frac{0.5}{3}[2 + 4(5.5) + 2(12) + 4(23) + 2(40) + 64.5] = \frac{0.5}{3}[2+22+24+92+80+64.5] = \frac{0.5}{3}[284.5] = \frac{142.25}{3} = 47.416666...$.

Using Simpson's rule:

$\frac{h}{3}[f(0) + 4f(0.5) + 2f(1) + 4f(1.5) + 2f(2) + f(2.5)]$

$= \frac{0.5}{3} [2 + 4(5.5) + 2(12) + 4(23) + 2(40) + 64.5] = \frac{0.5}{3}[2+22+24+92+80+64.5] = \frac{0.5}{3}(284.5) = 47.41666...$

However, we were told to use 5 segments, so n=5, and we need 6 values in Simpson's rule. Therefore, we have two intervals of length 2.5/2 = 1.25 with f(0) = 2, f(1.25) = 16.84375, and f(2.5)=64.

5.

Then Simpson's gives (1.25/3) [2 + 4(16.84375) + 64.5] = (1.25/3) [2 + 67.375 + 64.5] = (1.25/3) [133.875] = 55.78125

Relative error = $\frac{|55.78125 - 55.78125|}{55.78125} = 0$

f. $z = 10-4x^2-y^2$.

$\frac{\partial z}{\partial x} = -8x$ and $\frac{\partial z}{\partial y} = -2y$.

At $(1,2)$, $z = 10-4(1)^2-(2)^2 = 10-4-4 = 2$.

The trace of $z$ in the $xz$-plane is obtained by setting $y=2$: $z = 10 - 4x^2 - 4 = 6 - 4x^2$, $\frac{dz}{dx} = -8x$, so at $x=1$, the slope is $-8(1) = -8$.

The trace of $z$ in the $yz$-plane is obtained by setting $x=1$: $z = 10 - 4(1)^2 - y^2 = 6 - y^2$, $\frac{dz}{dy} = -2y$, so at $y=2$, the slope is $-2(2) = -4$.

g. $\sum_{n=1}^\infty \frac{n(-1)^n}{4^n} (x+3)^n$.

We use the Ratio Test to find the radius of convergence $R$.

$\lim_{n\to\infty} |\frac{a_{n+1}}{a_n}| = \lim_{n\to\infty} |\frac{\frac{(n+1)(-1)^{n+1}}{4^{n+1}} (x+3)^{n+1}}{\frac{n(-1)^n}{4^n} (x+3)^n}| = \lim_{n\to\infty} |\frac{(n+1)(x+3)}{4n}| = \frac{|x+3|}{4} \lim_{n\to\infty} \frac{n+1}{n} = \frac{|x+3|}{4}$.

For convergence, we require $\frac{|x+3|}{4} < 1$, so $|x+3| < 4$. Thus, the radius of convergence $R = 4$.

3. Final Answer

a. $z = \frac{1}{2}x + \frac{1}{4}y - 1 + \ln(4)$

b. The series is a p-series. It converges if $p > 1$ and diverges if $0 < p \le 1$.

c. $\frac{2}{\sqrt{x^2+y^2+z^2}}$

d. $\sum_{n=1}^\infty \frac{2^n}{n^3+1}$ diverges.

e. [i] $55.234375$, relative error is approximately $0.9803\%$. [ii] 55.78125, the relative error is

0. f. The slopes are -8 and -

4. g. 4