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202211041009 Solution to 2005-CE-AMATH-II-10

(a) Show that

\displaystyle{\frac{\mathrm{d}}{\mathrm{d}x}[x(x+1)^n]=(x+1)^{n-1}[(n+1)x+1]},

where n is a rational number.
(b) The slope at any point (x,y) of a curve C is given by

\displaystyle{\frac{\mathrm{d}y}{\mathrm{d}x}=(x+1)^{2004}(2006x+1)}.

If C passes through the point (-1,1), find the equation of C.

Roughwork.

(a)

Beginning by integration on the right hand side,

\begin{aligned} &\quad\, \int (x+1)^{n-1}[(n+1)x+1]\,\mathrm{d}x \\ & = \int (x+1)^{n-1}[(n+1)(x+1)-n]\,\mathrm{d}x \\ & = (n+1)\int (x+1)^n\,\mathrm{d}x - n\int (x+1)^{n-1}\,\mathrm{d}x \\ & = (n+1)\frac{(x+1)^{n+1}}{n+1} - n\frac{(x+1)^{(n-1)+1}}{(n-1)+1} + C \\ & = (x+1)^{n+1}-(x+1)^n + C \\ & = (x+1)^n(x+1) - (x+1)^n + C \\ & = x(x+1)^n + C \textrm{ for some constant}\\ \end{aligned}

or by differentiation on the left hand side,

\begin{aligned} &\quad\,\frac{\mathrm{d}}{\mathrm{d}x}[x(x+1)^n] \\ & = x\cdot \frac{\mathrm{d}}{\mathrm{d}x}\big( (x+1)^n\big) + \frac{\mathrm{d}}{\mathrm{d}x} (x)\cdot (x+1)^n \\ & = x\cdot (n)(x+1)^{n-1}(1) + 1\cdot (x+1)^n \\ & = (nx)(x+1)^{n-1}+(x+1)(x+1)^{n-1} \\ & = (x+1)^{n-1}[(n+1)x+1] \\ \end{aligned}

(b)

\begin{aligned} y & = \int \frac{\mathrm{d}y}{\mathrm{d}x}\,\mathrm{d}x \\ & = \int (x+1)^{(2005)-1}[( (2005)+1)x+1] \\ y(x) & \stackrel{\textrm{(a)}}{=} x(x+1)^{2005}+C\textrm{ for some constant} \\ \end{aligned}

Substituting -1 for x and 1 for y(x),

\begin{aligned} 1 & = (-1)((-1)+1)^{2005} + C \\ C & = 1 \\ \end{aligned}

In conclusion, the equation of C is

y=x(x+1)^{2005}+1.