Practice section 2 of 6

(Self-reading)

A cannonball is fired from O at speed v metres per second and angle $\theta^\circ$ above the ground. At time t seconds, $x=(v\cos\theta)t$, $y=(v\sin\theta)t-5t^2$, where x and y are horizontal and vertical displacement in metres.

(i) Find its horizontal range d when it hits the ground, in terms of v and theta.

Use $v=200$ for the remaining parts. A target approaches O along the ground at 10 metres per second, initially 3000 metres away when the cannonball is fired. Neglect the target height.

(ii) Show the possible firing angles to hit it are $22.7^\circ$ and $69.5^\circ$.

(iii) Explain which angle hits it earlier.

For $x=(1+t)^2$, $y=2t$, find $dy/dx$ in terms of t and hence the point where the gradient is 1.

Find $dy/dx$ in terms of t for (a) $x=4t$, $y=1-1/t$; (b) $x=4\cos t$, $y=3\sin t$.

For $x=a(1+1/t)$, $y=a(t-1/t^2)$, where a is constant and $t\ne0$, (i) find the normal at $t=2$; (ii) find the tangent at $(0,-2a)$.

For $x=t^2$, $y=2/t$:

(i) Find and simplify the tangent equation at $(p^2,2/p)$.

(ii) Hence find Q and R where that tangent meets the x- and y-axes, respectively.

(iii) Find a Cartesian equation for the locus of the midpoint of QR as p varies.

For $x=4t/(1-t^2)$, $y=1/(1-t^2)$, show the normal at $P(8/3,4/3)$ is $3y=-15x+44$. It meets the curve again at Q; find t at Q.