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Kieran C.

Master's graduate in mathematics and physics, Maths Olympiad contestant

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Linear Algebra

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Question:

The following quadratic equation $(x^{2}+ux+v=0$) has two roots, $$\alpha$$ and $$\beta$$. Derive an expression for $$\alpha^{2}+\beta^{2}$$ in terms of $$u$$ and $$v$$.

Kieran C.

Answer:

Ignore the temptation to apply the quadratic formula; there is an easier way to do this problem. As $$\alpha$$ and $$\beta$$ are roots of the quadratic equation: $(x^{2}+ux+v=(x-\alpha)(x-\beta)=0 \\ \Rightarrow x^{2}+ux+v=x^{2}-(\alpha+\beta)x+\alpha\beta$) Equating the coefficients of these two quadratic equations, we find that $$u=-(\alpha+\beta)$$ and $$v=\alpha\beta$$. To get $$\alpha^{2}+\beta^{2}$$, we square $$\alpha+\beta$$ and hope terms can be tidied up. $(u^{2}=(\alpha+\beta)^{2}=\alpha^{2}+\beta^{2}+2\alpha\beta=\alpha^{2}+\beta^{2}+2v$) So finally, $(\alpha^{2}+\beta^{2}=u^{2}-2v$)

Calculus

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Question:

Calculate the derivative of $(f(r)=\sqrt{\frac{2(R-r)}{r^{2}+2(R-r)^{2}}}$) at $r=0$.

Kieran C.

Answer:

We will split this problem into steps. Let $(g(r)=\frac{2(R-r)}{r^{2}+2(R-r)^{2}}=\frac{u}{v}$) $$f(r)=\sqrt{g(r)}$$, so by the chain rule $(f'(0)=\frac{g'(0)}{2\sqrt{g(0)}}=\frac{\sqrt{R}}{2}g'(0)$) $$u'(r)=-2$$ and $$v'(r)=2r-4(R-r)=6r-4r$$. Hence $$u(0)=2R$$, $$v(0)=2R^{2}$$, $$u'(0)=-2$$ and finally $$v'(0)=-4R$$. Hence by the quotient rule: $(g'(0)=\frac{v(0)u'(0)-u(0)v'(0)}{v(0)^{2}}=\frac{-4R^{2}+8R^{2}}{4R^{4}}=\frac{1}{R^{2}}$) So finally, $(f'(0)=\frac{\sqrt{R}}{2R^{2}}=\frac{1}{2\sqrt{R^{\,3}}}$) Note that initially, this problem looks very difficult, as you may be tempted to calculate $$f'(r)$$ and evaluate this at $$r$$. However, this is quite tricky, and not necessary at all.

Physics

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Question:

A ball is launched upwards with velocity $$u$$ from a height $$h$$ above the ground. How long will the ball be airborne before it hits the ground?

Kieran C.

Answer:

Denote the mass of the ball by $$m$$ and the acceleration due to gravity as $$g$$. The ball initially travels upwards until it stops at it's highest point, and then descends to the ground. Let the height of this highest point by $$H$$. We will work out the time it takes the ball to get from $$h$$ to $$H$$, and then from $$H$$ to the ground. $$h$$ to $$H$$: At the highest point, the velocity $$v$$ is $$0$$. Let the time taken be $$T_{1}$$. The initial velocity is $$u$$ and the acceleration is $$g$$ downwards. We know from the "uvast" equations that acceleration is equal to the difference in velocities divided by the time taken, so $(g=\frac{u}{T_{1}}\Rightarrow T_{1}=\frac{u}{g}$) The distance travelled $$s$$ during this leg of the journey can be calculated using $(v^{2}=u^{2}+2as\Rightarrow 0=u^{2}-2gs\Rightarrow s=\frac{u^{2}}{2g}$) Hence, the highest point reached by the ball is at height $(H=h+s=h+\frac{u^{2}}{2g}$) $$H$$ to $$0$$: Denote the travel time by $$T_{2}$$. The initial velocity is 0, the distance travelled is $$H$$. Then we use the following "uvast" equation to find $$T_{2}$$: $(s=ut+\frac{1}{2}at^{2}\Rightarrow H = \frac{1}{2}gT_{2}^{2}\Rightarrow T_{2}=\sqrt{\frac{2H}{g}}$) Finally, combining all these equations we get the total travel time to be $(T=T_{1}+T_{2}=\frac{u}{g}+\sqrt{\frac{2H}{g}}=\frac{u}{g}+\sqrt{\frac{2h}{g}+\left(\frac{u}{g}\right)^{2}}$)

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