QUESTION 1 (a) - GUIDED SOLUTION

Question (a) (i)

LaTeX: x=\frac{t}{1+t}

LaTeX: \frac{dx}{dt}=\frac{1\left(1+t\right)-t\left(1\right)}{\left(1+t\right)^2}=\frac{1}{\left(1+t\right)^2}

LaTeX: y=\frac{t^3}{1+t}

LaTeX: \frac{dy}{dt}=\frac{3t^2\left(1+t\right)-t^3\left(1\right)}{\left(1+t\right)^2}=\frac{3t^2+2t^3}{\left(1+t\right)^2}=\frac{t^2\left(3+2t\right)}{\left(1+t\right)^2}

LaTeX: \frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}=\frac{t^2\left(3+2t\right)}{\left(1+t\right)^2}\times\left(1+t\right)^2=t^2\left(3+2t\right)

At LaTeX: \left(\frac{1}{2},\:\frac{1}{2}\right)

LaTeX: x=\frac{t}{1+t}

LaTeX: \frac{1}{2}=\frac{t}{1+t}

LaTeX: 1+t=2t

LaTeX: t=1

LaTeX: \to\frac{dy}{dx}=1^2\left(3+2\left(1\right)\right)=5

Question (a) (ii)

LaTeX: y=mx+c using LaTeX: x=\frac{1}{2}, LaTeX: y=\frac{1}{2} and LaTeX: m=5

LaTeX: \frac{1}{2}=5\left(\frac{1}{2}\right)+c

LaTeX: -2=c

LaTeX: y=5x-2

when LaTeX: x=0

LaTeX: y=-2

LaTeX: y - intercept is LaTeX: \left(0,\:-2\right)

when LaTeX: y=0

LaTeX: x=\frac{2}{5}

LaTeX: x - intercept is LaTeX: \left(\frac{2}{5},\:0\right)