1. ## Implicit Differentiation

Hi

Could someone please show me how to do question 12?

2. Originally Posted by xwrathbringerx
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.

3. Hmm how exactly do I know that the tangents are horizontal and vertical?

Also could you please show me how to do (b) and (c)...I have no idea how to do them

4. 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 $\displaystyle x^3+ y^3= (x+ y)(x^2- xy+ y^2)= (x+y)(xy)(x/y- 1+ y/x)$.

5. hmmm for (b), when x = infinity, shouldnt y/x = infinity too? (since y/x = x/3 + y^2/3x^2?)

6. Originally Posted by xwrathbringerx
hmmm for (b), when x = infinity, shouldnt y/x = infinity too? (since y/x = x/3 + y^2/3x^2?)
Dear xwrathbringerx,

How did you get, $\displaystyle \frac{y}{x}=\frac{x}{3}+\frac{y^2}{3x^2}~??$

This is obviously a wrong equation since, when multiplied by $\displaystyle 3x^2$ you get, $\displaystyle 3xy=x^3+y^2$ which contradicts the original equation.

7. the 1st part of (b) says "show ...". I cross multiplied this to obtain it with y/x =

8. Originally Posted by xwrathbringerx
the 1st part of (b) says "show ...". I cross multiplied this to obtain it with y/x =
Dear xwrathbringerx,

Seems your cross multiplication is wrong. If you cross multiply you should obtain, $\displaystyle \frac{y}{x}=\frac{x}{3}+\frac{y^3}{3x^2}$

9. Ooops

Bur Sudharaka, doesnt that still give infinity?

10. Originally Posted by xwrathbringerx
Ooops

Bur Sudharaka, doesnt that still give infinity?
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??

11. OH

But I dont get it.Why did the question say y/x --> -1?

12. Originally Posted by xwrathbringerx
OH

But I dont get it.Why did the question say y/x --> -1?
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.