Implicit Differentiation

**xwrathbringerx** · April 3rd 2010, 12:57 AM

Hi

Could someone please show me how to do question 12?

**Sudharaka** · April 3rd 2010, 01:19 AM

Originally Posted by xwrathbringerx

Hi

Could someone please show me how to do question 12?

Dear xwrathbringerx,

By differentiation with respect to x,

$3x^2+3y^{2}\frac{dy}{dx}=3\left[y+x\frac{dy}{dx}\right]$

$[3y^{2}-3x]\frac{dy}{dx}=3y-3x^2$

$[y^2-x]\frac{dy}{dx}=y-x^2$

$\frac{dy}{dx}=\frac{y-x^2}{y^2-x}$

Can you do it from here?? If you need more help don't hesitate to ask.

Hope this helps you.

**xwrathbringerx** · April 3rd 2010, 03:29 AM

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

**HallsofIvy** · April 3rd 2010, 04:47 AM

A horizontal line has slope 0, a vertical line does not have a slope.

For (b) you are to show that $\frac{1+ (y/x)^3}{y/x}= 3/x$ .

Starting from the original equation, $x^3+ y^3= 3xy$ , divide both sides by $x^3$ to get $1+ (y/x)^3= 3y/x^2= 3(y/x)/x$ . Now multiply both sides by y/x.

For (c) you need to know that $x^3+ y^3= (x+ y)(x^2- xy+ y^2)= (x+y)(xy)(x/y- 1+ y/x)$ .

**xwrathbringerx** · April 3rd 2010, 05:03 AM

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

**Sudharaka** · April 3rd 2010, 05:43 AM

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, $\frac{y}{x}=\frac{x}{3}+\frac{y^2}{3x^2}~??$

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

**xwrathbringerx** · April 3rd 2010, 05:46 AM

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

**Sudharaka** · April 3rd 2010, 05:53 AM

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, $\frac{y}{x}=\frac{x}{3}+\frac{y^3}{3x^2}$

**xwrathbringerx** · April 3rd 2010, 05:57 AM

Ooops

Bur Sudharaka, doesnt that still give infinity?

**Sudharaka** · April 3rd 2010, 06:09 AM

Originally Posted by xwrathbringerx

Ooops

Bur Sudharaka, doesnt that still give infinity?

Dear xwrathbringer,

Suppose when $x\rightarrow{\infty}\Rightarrow{y\rightarrow{-\infty}}$ . Therefore generally we cannot say that $\frac{y}{x}\rightarrow{\infty}$ since we don't know what happens to y when x goes to infinity. Does this solve your problem??

**xwrathbringerx** · April 3rd 2010, 06:16 AM

OH

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

**Sudharaka** · April 3rd 2010, 06:24 AM

Originally Posted by xwrathbringerx

OH

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

Now consider, $\frac{1+\left(\frac{y}{x}\right)^{3}}{\frac{y}{x}} =\frac{3}{x}$

When, $\frac{y}{x}\rightarrow{-1}\Rightarrow{}\frac{1+\left(\frac{y}{x}\right)^{3 }}{\frac{y}{x}}\rightarrow{0}\Rightarrow{\frac{3}{ x}\rightarrow{0}\Rightarrow{x\rightarrow{\infty}}}$

Get the idea??? If you have any questions please don't hesitate to ask. I'll try my best to answer them.

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