Answer to Question #329435 in Calculus for Rohit

Question

trace the curve , x^2= y^2((x+1)^3 ) , stating all the points used in process

Answer 1

ANSWER

Let $\Gamma$ be the curve defined by the equation $x^2=y^2(x+1)^3$ .

$\Gamma=\left \{ \left ( x,y \right ):x>-1, x^{2} =y^{2}\left ( x+1 \right )^{3}\right \}$ ,

then $\Gamma=\Gamma_{1}\cup\Gamma_{2}$ , where $\Gamma_{1}=$$\left \{ \left ( x,f(x)) \right ):x>-1 \right \}$ , $\Gamma_{2}= \left \{ \left ( x,-f(x)) \right ):x>-1 \right \},$ $f(x)= \frac{x}{(x+1)^{\frac{3}{2}}}$ . $f(0)=0$

$f'(x)=\frac { { \left( x+1 \right) }^{ \frac { 3 }{ 2 } }-\frac { 3 }{ 2 } x{ \left( x+1 \right) }^{ \frac { 1 }{ 2 } } }{ { \left( x+1 \right) }^{ 3\quad } }$ $=\frac{1}{2}\cdot \frac{(2-x)}{(1+x)^\frac{5}{2}}$

$f''(x)=-\frac { 1 }{ 2 } \cdot \frac { { \quad \left( x+1 \right) }^{ \frac { 5 }{ 2 } }+\frac { 5 }{ 2 } (2-x){ \left( x+1 \right) }^{ \frac { 3 }{ 2 } } }{ { \left( x+1 \right) }^{ 5\quad } } =\frac { 3 }{ 4 } \cdot \frac { (x-4) }{ { \left( x+1 \right) }^{ \frac { 7 }{ 2 } } }$

In the interval $(-1,2)$ $f'(x)>0$ , so the function $f$ increases. In the interval $( 2 ,+\infty)$ $f'(x)<0$ , so the function $f$ decreases. Thus $max f(x)=f(2)=$ $\frac{2}{3\sqrt{3}}$ .

On the graph of the function this is point B.

$f''(x)<0$ if $x\in(-1,4)$ and $f''(x)>0$ if $x\in(4, +\infty)$ .Therefore, the function $f$ in the interval $(-1,4)$ is convex upwards , in the interval $(4, +\infty)$ is convex downwards (see point D on the graph of the function).

$\lim _{ x\rightarrow +\infty }{ f(x)= } \lim _{ x\rightarrow +\infty }{ \frac { x }{ { x }^{ \frac { 3 }{ 2 } }{ \left( 1+\frac { 1 }{ x } \right) }^{ \frac { 3 }{ 2 } } } = } \lim _{ x\rightarrow +\infty }{ \frac { 1 }{ { x }^{ \frac { 1 }{ 2 } }{ \left( 1+\frac { 1 }{ x } \right) }^{ \frac { 3 }{ 2 } } } = } 0$

$\lim _{ x\rightarrow -1 }{ f(x)= } \lim _{ x\rightarrow -1 } \frac { x }{ { \left( 1+x \right) }^{ \frac { 3 }{ 2 } } } { =-\infty }$ . Hence, $y=0$ is the horizontal asymptote, $x=-1$ is the vertical asymptote.