November 23, 2022 – Physicspupil

November 23, 2022November 23, 2022

202211231608 Solution to 1982-CE-AMATH-I-5

Show that the tangents to the curve

$y=x^3-9x^2+30x+4$

cannot be parallel to the $x$ -axis.

Roughwork.

The slope $y'(x)$ of tangent to the curve is

$\begin{aligned} y'(x) & = \frac{\mathrm{d}y}{\mathrm{d}x} \\ & = \frac{\mathrm{d}}{\mathrm{d}x}(x^3-9x^2+30x+4) \\ & = 3x^2-18x+30 \\ \end{aligned}$

Assume to the contrary that such tangent as $y'(x)=0$ exists. Then

$\begin{aligned} 3x^2-18x+30 & = 0 \\ \frac{-(-18)\pm\sqrt{(-18)^2-4(3)(30)}}{2(3)} & = x \\ \end{aligned}$

where the discriminant

$\Delta = -36<0$ ,

implies there are no real roots. Contradiction, which was to be demonstrated.

November 23, 2022November 23, 2022

202211231536 Solution to 1986-CE-AMATH-I-4

Find the equation of the tangent to the curve

$x^2+xy+y^2=7$

at the point $(2,1)$ .

Roughwork.

$\begin{aligned} x^2+xy+y^2-7 & = 0 \\ 2x\,\mathrm{d}x+x\,\mathrm{d}y+y\,\mathrm{d}x + 2y\,\mathrm{d}y & = 0 \\ 2x + x\frac{\mathrm{d}y}{\mathrm{d}x}+y+2y\frac{\mathrm{d}y}{\mathrm{d}x} & = 0 \\ -\frac{2x+y}{x+2y} & = y'(x) \\ \end{aligned}$

The slope of tangent at $(x=2,y=1)$ is

$y'(x)|_{(2,1)}=\displaystyle{-\frac{2(2)+(1)}{(2)+2(1)}=-\frac{5}{4}}$ .

and its equation

$\begin{aligned} & \quad\enspace \frac{y-1}{x-2} = -\frac{5}{4} \\ & \Rightarrow 5x + 4y = 14 \\ \end{aligned}$

November 23, 2022November 23, 2022

202211231154 Solution to 1969-HL-PHY-I-2

An aeroplane flying with a uniform speed $v$ in a small circle of radius $r$ at a constant altitude $h$ releases a body. Assuming that air resistance can be neglected, find the angle of inclination between the ground and the line of sight of the pilot when he observes the body touching the ground.