Hi
Could someone please show me how to do question 12?
Dear xwrathbringerx,
By differentiation with respect to x,
$\displaystyle 3x^2+3y^{2}\frac{dy}{dx}=3\left[y+x\frac{dy}{dx}\right]$
$\displaystyle [3y^{2}-3x]\frac{dy}{dx}=3y-3x^2$
$\displaystyle [y^2-x]\frac{dy}{dx}=y-x^2$
$\displaystyle \frac{dy}{dx}=\frac{y-x^2}{y^2-x}$
Can you do it from here?? If you need more help don't hesitate to ask.
Hope this helps you.
A horizontal line has slope 0, a vertical line does not have a slope.
For (b) you are to show that $\displaystyle \frac{1+ (y/x)^3}{y/x}= 3/x$.
Starting from the original equation, $\displaystyle x^3+ y^3= 3xy$, divide both sides by $\displaystyle x^3$ to get $\displaystyle 1+ (y/x)^3= 3y/x^2= 3(y/x)/x$. Now multiply both sides by y/x.
For (c) you need to know that $\displaystyle x^3+ y^3= (x+ y)(x^2- xy+ y^2)= (x+y)(xy)(x/y- 1+ y/x)$.
Dear xwrathbringer,
Suppose when $\displaystyle x\rightarrow{\infty}\Rightarrow{y\rightarrow{-\infty}}$. Therefore generally we cannot say that $\displaystyle \frac{y}{x}\rightarrow{\infty}$ since we don't know what happens to y when x goes to infinity. Does this solve your problem??
Now consider, $\displaystyle \frac{1+\left(\frac{y}{x}\right)^{3}}{\frac{y}{x}} =\frac{3}{x}$
When, $\displaystyle \frac{y}{x}\rightarrow{-1}\Rightarrow{}\frac{1+\left(\frac{y}{x}\right)^{3 }}{\frac{y}{x}}\rightarrow{0}\Rightarrow{\frac{3}{ x}\rightarrow{0}\Rightarrow{x\rightarrow{\infty}}}$
Get the idea??? If you have any questions please don't hesitate to ask. I'll try my best to answer them.