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# Tutor profile: Krista K.

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Krista K.
Tutor for 8 years in calculus, current neuroscience graduate student
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## Questions

### Subject:Linear Algebra

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Question:

Find the eigenvalues and eigenvectors for the matrix $$A = \begin{bmatrix} 2 & -1\\ -1 & 2\end{bmatrix}$$

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Krista K.

The eigenvalues of $$A$$ come from $$det(A-\lambda I) =0$$ Remember that $$\lambda I$$ is equivalent to $$\begin{bmatrix} \lambda & 0\\ 0 & \lambda\end{bmatrix}$$ Then we can see $$A - \lambda I = \begin{bmatrix} 2 & -1\\ -1 & 2\end{bmatrix}-\begin{bmatrix} \lambda & 0\\ 0 & \lambda\end{bmatrix}$$ Recall that matrix subtraction (and addition) is done element by element. So then we get: $$A- \lambda I = \begin{bmatrix} 2-\lambda && -1 \\ -1 && 2-\lambda \end{bmatrix}$$ We need the determinant, which is written as follows: $$det(A-\lambda I) = \begin{vmatrix} 2-\lambda & -1\\ -1 & 2-\lambda\end{vmatrix}$$ In this case, the determinant is: $$(2-\lambda)(2-\lambda)-(-1)(-1)$$ We can simplify this: $$4- 4\lambda + \lambda^2 + 1$$ Further simplification: $$5 - 4\lambda + \lambda^2$$ Next, to get the eigenvalues, we need to find the roots of this polynomial. This requires factoring. $$5 - 4\lambda + \lambda^2 = (\lambda-1)(\lambda-3)$$ Therefore, the roots are 1 and 3. The eigenvalues are 1 and 3. Next, we find the eigenvectors. The equation for eigenvectors is $$(A -\lambda I) \mathbf{x} = 0$$ For $$\lambda = 1$$: $$(A -\lambda I) \mathbf{x} = \left ( \begin{bmatrix} 2 && -1 \\ -1 && 2 \end{bmatrix} - 1* \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} \right ) \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ We can simplify this: $$\begin{bmatrix} 2-1 && -1 \\ -1 && 2-1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} 1 && -1 \\ -1 && 1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} =\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$ Using matrix multiplication, we get: $$x-y=0$$ and $$y-x=0$$. Solving this system of equations gives us $$x=y$$ so $$x=y=1$$. For our eigenvalue, $$\lambda_1 = 1$$ the eigenvector is $$x_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ Now we repeat the above steps for $$\lambda = 3$$: $$(A -\lambda I) \mathbf{x} = \left ( \begin{bmatrix} 2 && -1 \\ -1 && 2 \end{bmatrix} - 3* \begin{bmatrix} 1 && 0 \\ 0 && 1 \end{bmatrix} \right ) \begin{bmatrix} x \\ y \end{bmatrix}$$ We can simplify this: $$\begin{bmatrix} 2-3 && -1 \\ -1 && 2-3 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} = \begin{bmatrix} -1 && -1 \\ -1 && -1\end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix}$$ Using matrix multiplication, we get: $$-x-y=0$$ and $$-y-x=0$$. Solving this system of equations gives us $$x=-y$$ so $$x=1$$ and $$y = -1$$ For our eigenvalue, $$\lambda_2 = 3$$ the eigenvector is $$x_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$ Final Solution: Eigenvalue $$\lambda_1 = 1$$ has eigenvector $$x_1 = \begin{bmatrix} 1 \\ 1 \end{bmatrix}$$ and eigenvalue $$\lambda_2 = 3$$ has eigenvector $$x_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix}$$

### Subject:Pre-Calculus

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Question:

Use trig identities to simplify: $$sin^4\theta \ cos^3 \theta$$ to $$(sin^4 \theta-sin^5 \theta) \ cos\theta$$

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Krista K.

First we start by simplifying the cosine in order to use a trig identity: $$sin^4\theta \ (cos^2 \theta) \ cos\theta$$ Then we substitute the trig identity: $$cos^2 \theta = 1-sin^2\theta$$ $$sin^4\theta \ (1-sin^2\theta) \ cos\theta$$ Next, we simplify by distributing $$sin^4\theta$$: $$(sin^4\theta*1-sin^4\theta *sin^2\theta) \ cos\theta$$ Lastly, we finish simplifying: $$(sin^4\theta*1-sin^6\theta) \ cos\theta$$

### Subject:Calculus

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Question:

The curve $$y=\sqrt{4-x^2},-1\leqslant x \leqslant 1$$ is an arc of the circle $$x^2 + y^2 = 4$$. Find the area of the surface obtained by rotating this arc about the x-axis.

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Krista K.

We know that the equation for the surface area is: $$S = \int_{-1}^{1}2\pi y \sqrt{1+(\frac{dy}{dx})^2} \ dx$$ So, we start by taking the derivative of y with respect to x. $$\frac{dy}{dx} = \frac{1}{2}(4-x^2)^{-\frac{1}{2}}(-2x)=\frac{-x}{\sqrt{4-x^2}}$$ We then insert this derivative and the original equation (y) into the above question: $$\int_{-1}^{1}2\pi \sqrt{4-x^2} \sqrt{1+ \left ( \frac{-x}{\sqrt{4-x^2}} \right )^2} \ dx$$ Then we can start simplifying, starting with the exponent inside of the square root: $$\int_{-1}^{1}2\pi \sqrt{4-x^2} \sqrt{1+ \frac{x^2}{4-x^2}} \ dx$$ Next, we simplify further by combining fractions and redistributing the square root: $$\int_{-1}^{1}2\pi \sqrt{4-x^2} \frac{2}{\sqrt{4-x^2}} \ dx$$ Now, you'll notice we can pull the $$2\pi$$ as well as the $$2$$ to the front of the integral: $$2*2\pi \int_{-1}^{1} \sqrt{4-x^2} \frac{1}{\sqrt{4-x^2}} \ dx$$ The integral then simplifies further to: $$4\pi \int_{-1}^{1} 1 \ dx$$ Solving the integral, we get: $$4\pi x \bigg\rvert_{-1}^{1}$$ Evaluating the function, we have: $$4\pi * (1) - 4\pi *(-1)$$ $$4\pi + 4\pi$$ $$8\pi$$ The final solution is $$8\pi$$.

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