Practice section 5 of 6

A fish population satisfies $dP/dt=0.02P(100-P)$, with $t$ in weeks.

(i) Find the range of $P$ for strictly increasing population. Initially $P=20$.

(ii) Show that $P=100/(1+4e^{-2t})$.

(iii) Describe the long-run population and sketch $P$ against $t$.

In a loan scheme, the amount owed is $M$ dollars at time $t$ months; both variables are continuous. Interest increases the amount at a rate proportional to the amount owed, while repayment occurs at a constant rate of $500 per month. Interest and repayment balance when $M=10000$. An initial loan of $8000 is taken on 1 March 2013.

(i) Show $dM/dt=(M-10000)/20$. [2]

(ii) Find $M$ in terms of $t$. [4]

(iii) Find the date when the loan is cleared. [2]

Show that the general solution of $y(dy/dx)=2x-1$, $y\ge0$, is $y^2=2x^2-2x+C$, where $C$ is arbitrary. Sketch the curve passing through $(0,1)$.

(i) Use $z=(2x-1)dy/dx$ to express $(2x-1)d^2y/dx^2=2(1-dy/dx)$ as an equation in $z,x$. [2]

(ii) Solve with $y=2$ and $dy/dx=2$ at $x=0$. [5]

(iii) Describe the gradient as (a) $x\to\pm\infty$ [1] and (b) $x\to1/2$. [1]

(iv) Sketch the solution from (ii). [2]

Using $y=vx$, solve $dy/dx=y(x+y)/[x(y-x)]$.

(a) Find $y=f(x)$ through $(1,0)$ and $(-1,2)$ with $dy/dx=3x^2-x^{-2}+C$. [3]

(b) Perry jogs a distance $x$ kilometres in $t$ hours. The difference between his initial speed and his speed at time $t$ is proportional to distance covered. He starts at $5$ km/h and travels at $4$ km/h after $2$ km. Find the exact time for a $5$ km jog. [6]