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Tutor profile: Conner G.

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Conner G.
Graduate student in mathematics
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Questions

Subject: Linear Algebra

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Question:

Use the Gram-Schmidt process to find an orthogonal basis for $$W=\operatorname{span}\{\begin{pmatrix}2 \\ 0 \\ 1\end{pmatrix},\begin{pmatrix}1\\3\\0\end{pmatrix}\}.$$

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Conner G.
Answer:

First note that our two vectors are linearly independent. Indeed, two vectors being linearly dependent is equivalent to saying that they are scalar multiples of each other. This is not the case as $$\lambda\begin{pmatrix}2 \\ 0 \\ 1\end{pmatrix}=\begin{pmatrix}1\\3\\0\end{pmatrix}$$ implies that $$2\lambda=1.$$ But then $$\lambda0=0\ne 3.$$ Now, we apply the Gram-Schmidt process. We start by taking $$v_1=\begin{pmatrix}2 \\ 0 \\ 1\end{pmatrix}.$$ We now need a vector in $$W$$ orthogonal to $$v_1.$$ The difference between $$\begin{pmatrix}1\\3\\0\end{pmatrix}$$ and its orthogonal projection onto the space spanned by $$v_1$$ will give us exactly such a vector. So we have $(v_2= \begin{pmatrix}1\\3\\0\end{pmatrix} -\frac{v_1 \cdot \begin{pmatrix}1\\3\\0\end{pmatrix}}{v_1 \cdot v_1}\begin{pmatrix}1\\3\\0\end{pmatrix} = \begin{pmatrix}1\\3\\0\end{pmatrix} - \frac{2}{5}\begin{pmatrix}1\\3\\0\end{pmatrix} = \begin{pmatrix}3/5\\9/5\\0\end{pmatrix}.$) We have $$\{\begin{pmatrix}2 \\ 0 \\ 1\end{pmatrix},\begin{pmatrix}3/5\\9/5\\0\end{pmatrix}\}$$ being an orthogonal basis for $$W.$$

Subject: Calculus

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Question:

Use implicit differentiation to find an equation of the tangent line to the solutions of $(\sin\left(x+y\right) = 2x-2y$)at the point $$\left(\pi,\pi\right)$$.

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Conner G.
Answer:

Note that our given equation cannot be expressed as either a function in $$x$$ or a function in $$y.$$ This means we have to use implicit differentiation in order to get the slope of the tangent line at the given point. We first treat $$y$$ as a function of $$x$$ and then differentiate both sides with respect to $$x,$$ $(\frac{d}{dx}\sin\left(x+y\right)= \frac{d}{dx}\left[2x-2y\right].$)This gives us, $(\cos\left(x+y\right)\left(1+y'\right)=2-2y'.$) We now solve for $$y',$$ $(y'\cos\left(x+y\right)+2y'=2-\cos\left(x+y\right)$)$(y'\left(\cos\left(x+y\right)+2\right) = 2-\cos\left(x+y\right)$)$(y'=\frac{2-\cos\left(x+y\right)}{2+\cos\left(x+y\right)}$)Therefore, the slope of the tangent line at the point $$\left(\pi,\pi\right)$$ is $(y'\left(\pi,\pi\right) = \frac{2-\cos\left(2\pi\right)}{2+\cos\left(2\pi\right)} = \frac{2-1}{2+1}=\frac{1}{3}.$)We now use either the point slope formula or the $$y$$-intercept formula for the equation of a line, $(\pi=\frac{1}{3}\pi+b$)Therefore, $$b=\frac{2\pi}{3}$$ and we have the equation of our tangent line to the curve at the given point, $(y=\frac{1}{3}x+\frac{2\pi}{3}.$)

Subject: Differential Equations

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Question:

Solve $$x\frac{dy}{dx}+y=x^2y^2$$. Note: This is a classic example of a Bernoulli differential equation.

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Conner G.
Answer:

A differential equation of the form $(\frac{dy}{dx}+P\left(x\right)y=f\left(x\right)y^n,$)where $$n$$ is a real number, is called a Bernoulli differential equation. For $$n=0$$ or $$n=1$$ it is a linear equation. When $$n$$ is neither $$0$$ nor $$1,$$ we can use the substitution $$u=y^{1-n}$$ to get a linear equation. We can solve $(x\frac{dy}{dx}+y=x^2y^2$)using this technique. We first divide both sides by $$x$$ to get $(\frac{dy}{dx}+\frac{1}{x}y=xy^2.$)We now have our equation written in a form that exactly reflects our definition of a Bernouli equation. We have $$P\left(x\right) = \frac{1}{x},$$ $$f\left(x\right) = x,$$ and $$n=2.$$ We want to to choose the correct substitution which will be of the form $$u=y^{1-n}.$$ Now $$n=2,$$ so the substitution we want is $$u=y^{-1}.$$ To replace $$\frac{dy}{dx}$$ we differentiate $$y=u^{-1}$$ with respect to $$x.$$ (Remember, $$y$$ and $$u$$ are both functions in $$x$$.) $(\frac{dy}{dx}= \frac{d}{dx}u^{-1}=-u^{-2}\frac{du}{dx}$)As a result, we have $(-u^{-2}\frac{du}{dx}+\frac{1}{x}u^{-1}=xu^{-2}.$)Multiplying both sides by $$-u^2$$ gives us the linear equation $(\frac{du}{dx}-\frac{1}{x}u=-x,$)which has an integrating factor of $(e^{-\int\frac{dx}{x}} = e^{-\ln x} = x^{-1}$)on the interval $$\left(0,\infty\right).$$ We now multiply both sides by our integrating factor to get $(\frac{d}{dx}\left[x^{-1}u\right] =-1.$)Integrating both sides, $(x^{-1}u=-x+c.$)We now solve for $$u,$$ $(u=-x^2+cx$) and then substitute $$u=y^{-1}$$ $(\frac{1}{y} = -x^2+cx.$) Thus $(y=\frac{1}{-x^2+cx}$) is the solution to our differential equation.

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