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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 $$\vec{r}(t)=(\sin t, \cos t, t/6), 0 \leq t\leq 2\pi$$

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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.

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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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