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# Tutor profile: Samuel C.

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Samuel C.
Tutor for five years
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## Questions

### Subject:Linear Algebra

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Question:

Compute the eigenvalues and corresponding eigenvectors of the following matrix: $(\begin{bmatrix}-5&-3\\4&3\end{bmatrix}.$)

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Samuel C.

To find the eigenvalues of this matrix, we must find the determinant of $(\begin{bmatrix}-5-\lambda&-3\\4&3-\lambda\end{bmatrix},$) set it equal to 0, and solve for $$\lambda$$. We compute \begin{align} \det\Bigg(\begin{bmatrix}-5-\lambda&-3\\4&3-\lambda\end{bmatrix}\Bigg)&=(-5-\lambda)(3-\lambda)-(-3)(4)\\ &=-15+5\lambda-3\lambda+\lambda^2+12\\ &=\lambda^2+2\lambda-3\\ &=(\lambda-1)(\lambda+3)=0. \end{align} And find the eigenvalues to be $$-3$$ and $$1$$.

### Subject:Calculus

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Question:

Evaluate $$\int_C\vec{F}\cdot d\vec{r}$$ where $$\vec{F}(x,y,z)=(6z, y^2, 12x)$$ along $$\vec{r}(t)=(\sin t, \cos t, t/6), 0 \leq t\leq 2\pi$$

Inactive
Samuel C.

First, we compute $$\vec{r}'(t)=(\cos t,-\sin t,1/6)$$ and $$\vec{F}\big(\vec{r}(t)\big)=(t, \cos^2t, 12\sin t)$$. Then we have $(\int_C\vec{F}\cdot d\vec{r}=\int_0^{2\pi}t\cos t -\sin t\cos^2t+2\sin t\ dt.$) While the third term in the integral is simple to compute, the first must be done by parts and the second by substitution. We will first compute $$\int_0^{2\pi} t\cos t\ dt$$. We let $$u=t$$ and $$dv=\cos t\ dt$$. It follows that $$du = dt$$ and $$v=\sin t$$. Then we have \begin{align} \int_0^{2\pi} t\cos t\ dt & = t\sin t\ \Bigg|_0^{2\pi}-\int_0^{2\pi} \sin t\ dt \\ & = t\sin t + \cos t\ \Bigg|_0^{2\pi}\\ & = 2\pi \sin(2\pi)+\cos(2\pi)-\big(0\sin(0)+\cos(0)\big)\\ & = 2\pi\cdot 0 +1-1\\ & =0. \end{align} Now we will evaluate the second term $$\int_0^{2\pi}-\sin t\cos^2t\ dt$$ by letting $$u=\cos t$$ and $$du=-\sin t\ dt.$$ We compute \begin{align} \int_0^{2\pi}-\sin t\cos^2t\ dt&=\int_{t=0}^{t=2\pi}u^2\ du\\ &=\frac{1}{3}u^3\Bigg|_{t=0}^{t=2\pi}\\ &=\frac{1}{3}\cos^3 t\Bigg|_{0}^{2\pi}\\ &=\frac{1}{3}\cos^3 (2\pi)-\frac{1}{3}\cos^3 (0)\\ &=\frac 1 3-\frac 1 3 \\ &= 0. \end{align} Finally, we have $(\int_0^{2\pi}2\sin t\ dt=-2\cos t\Bigg|_{0}^{2\pi}=-2\cos(2\pi)-\big(-2\cos(0)\big)=-2+2=0$) and so $$\int_C\vec{F}\cdot d\vec{r}=0$$.

### Subject:Differential Equations

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Question:

Solve the initial value problem $$y''+4y'+13y=3u(t-4), y(0)=1, y'(0)=0$$ where $$u(t)$$ is the unit step function.

Inactive
Samuel C.

Given that this problem involves the unit step function, we will solve it using Laplace transforms. Applying the Laplace transform to both sides of the equation gives us $( s^2Y(s)-sy(0)-y'(0)+4(sY(s)-y(0))+13Y(s)=\frac{3}{s}e^{-4s}.$) We substitute $$y(0)=1$$ and $$y'(0)=0$$ and solve for Y(s) to obtain $(Y(s)=\frac{3}{s(s^2+4s+13)}e^{-4s}+\frac{s+4}{s^2+4s+13}.$) Using partial fraction decomposition we find $(Y(s)=\frac{3}{13}\bigg(\frac{1}{s}-\frac{s+4}{s^2+4s+13}\bigg)e^{-4s}+\frac{s+4}{s^2+4s+13}.$) We then complete the square on $$s^2+4s+13$$ to obtain $(Y(s)=\frac{3}{13}\bigg(\frac{1}{s}-\frac{s+4}{(s+2)^2+9}\bigg)e^{-4s}+\frac{s+4}{(s+2)^2+9}.$) Separating the terms further we get $(Y(s)=\frac{3}{13}\bigg(\frac{1}{s}-\frac{s+2}{(s+2)^2+9}-\frac{2}{(s+2)^2+9}\bigg)e^{-4s}+\frac{s+2}{(s+2)^2+9}+\frac{2}{(s+2)^2+9}.$) And we are finally able to compute the inverse Laplace transform, providing us with our solution: $(y=\frac{3}{13}\bigg(1-e^{-2(t-4)}\cos\big(3(t-4)\big)-\frac{2}{3}e^{-2(t-4)}\sin\big(3(t-4)\big)\bigg)u(t-4)+e^{-2t}\cos3t+\frac{2}{3}e^{-2t}\sin3t.$)

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