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Krista K.
Tutor for 8 years in calculus, current neuroscience graduate student
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Linear Algebra
TutorMe
Question:

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

Krista K.
Answer:

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 $$ x_2 = \begin{bmatrix} 1 \\ -1 \end{bmatrix} $$

Pre-Calculus
TutorMe
Question:

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

Krista K.
Answer:

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: $$ (sin^4\theta*1-sin^6\theta) \ cos\theta $$

Calculus
TutorMe
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.

Krista K.
Answer:

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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