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# Tutor profile: Chris S.

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Chris S.
Mathematics lecturer, postdoctoral fellow, PhD in mathematics in 2016
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## Questions

### Subject:Calculus

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Question:

Find the length of the curve $$y(x) = \sin^{-1}x + \sqrt{1-x^2}$$.

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Chris S.

Firstly, we realise that the domain of the given function is $$[-1,1]$$, and that \begin{equation*} y' = \frac{1}{\sqrt{1-x^2}} - \frac{x}{1-x^2} = \frac{1-x}{\sqrt{1-x^2}}. \end{equation*} Thus the length of the curve is given by \begin{align*} L &= \int_{-1}^1\sqrt{1 + \frac{(1-x)^2}{1-x^2}}\,\text{d}x \\ &= \int_{-1}^1\sqrt{1 + \frac{1-x}{1+x}}\,\text{d}x \\ &= \int_{-1}^1\sqrt{\frac{2}{1+x}}\,\text{d}x \\ &= 2\sqrt{2}\sqrt{1+x}\Bigm |_{-1}^1 \\ &= 4. \end{align*}

### Subject:Partial Differential Equations

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Question:

Given $$u_t + 25u_x = 0.$$, determine the characteristic coordinates for $$u=u(x,t)$$.

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Chris S.

Without loss of generality, we construct our characteristic curves from the initial curve $$t=0$$. The characteristic equations (where $$p$$ is the characteristic coordinate parametrising the characteristic curves and $$q$$ is the characteristic coordinate labeling the individual characteristic cureves) are $$\frac{dt}{dp} = 1,\quad\frac{dx}{dp}=25,$$ which after integration yield $$t = p + C_0,\quad x = 25p + C_1.$$ We fix the two degrees of freedom $$C_0,\,C_1$$ by requiring on $$t=0$$: $$p=0$$ and $$q=x$$. This yields $$C_0 = 0,\,C_1 = q$$ and we find the characteristic coordinates given by $$t = p,\quad x = 25p + q.$$ We are interested in the $$q=$$constant curves as these represent our characteristic curves. We have $$q = x - 25p = x - 25t.$$ Thus the characteristic curves are given by $$t = \frac{x}{25} + \;$$constant. These are just lines in the $$x-t$$ plane with slope $$1/25$$.

### Subject:Differential Equations

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Question:

Solve the initial value problem $$2y''(x)- 3 y(x) = 0,\quad y(0)=0,\;y(\pi/2)=1.$$

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Chris S.

Using the method of undetermined coefficients, we try $$y(x)=A\sin(x) + B\cos(x)$$ where $$A,B\in\mathbb{R}$$. This gives $$y'(x)=A\cos(x) - B\sin(x)$$ and $$y''(x)=-(A\sin(x) + B\cos(x))$$. We now substitute this into the differential equation to get $$2[-A\sin(x) - B\cos(x)] - [A\sin(x) + B\cos(x)] = 0\Leftrightarrow -3A\sin(x) - 3B\cos(x)=0$$. Now we impose the initial data, which yields $$y(0)=0\Rightarrow -3B=0 \Rightarrow B=0$$ and $$y(\pi/2)=1\Rightarrow -3A=1\Rightarrow A=-\frac13$$. Thus the unique solution to this differential equation with the given initial conditions is $$y(x)=-\frac13\sin(x)$$.

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