# Implicit Differentiation

• May 10th 2011, 08:48 PM
RezMan
Implicit Differentiation
Hey guys can you help me out with this:
Given the ellipse 4x^2 + 9y^2 = 36,
a) Find $\displaystyle \frac{dy}{dx}$.
is this correct:
$\displaystyle \frac{d(4x^2)}{dy}+\frac{d(9y^2)}{dx}*\frac{dy}{dx } = \frac{d(36)}{dx}$
$\displaystyle 8x+18y*\frac{dy}{dx}=0$
$\displaystyle \frac{dy}{dx}=-\frac{4x}{9y}$

b) Give the equations of all tangent lines to the graph at x = 1.

at this point I plug in 1 in the original function to get y right?

I got this weird number for y when I did that: $\displaystyle y=\pm \frac{4\sqrt{2}}{3}$

are my calculations correct?
• May 10th 2011, 08:57 PM
TheChaz
Quote:

Originally Posted by RezMan
Hey guys can you help me out with this:
Given the ellipse 4x^2 + 9y^2 = 36,
a) Find $\displaystyle \frac{dy}{dx}$.
is this correct:
$\displaystyle \frac{d(4x^2}{dy}+\frac{9y^2}{dx}*\frac{dy}{dx} = \frac{d(36)}{dx}$
$\displaystyle 8x+18y*\frac{dy}{dx}=0$
$\displaystyle \frac{dy}{dx}=-\frac{4x}{9y}$

b) Give the equations of all tangent lines to the graph at x = 1.

at this point I plug in x in the original function to get y right?

I got this weird number for y when I did that: $\displaystyle y=\pm \frac{4\sqrt{2}}{3}$

are my calculations correct?

a) yes
b) yes, but you're not done. Now you have points and can find slopes (dy/dx), so it's time to actually write down the equations!
• May 10th 2011, 08:59 PM
Prove It

I agree with your y value. Now you need to find the gradient of the tangent using your (x,y) coordinate.
• May 10th 2011, 09:08 PM
RezMan
so the point is actually (1,$\displaystyle \pm \frac{4\sqrt{2}}{3}$)?

so I'm correct in pluggin in 1 in the original function?

so
$\displaystyle -\frac{4(1)}{9(\frac{4\sqrt{2}}{3})}$

slope=$\displaystyle \pm \frac{1}{3\sqrt{2}}$

?
• May 10th 2011, 09:10 PM
Prove It
First of all, you need to work out each point separately...

No, don't plug each point into the original function. Plug each point into the derivative function to find the gradient of each tangent line.

Once you have the gradient, then you have x, y, m for the y = mx + c equation, solve for c. Then you have the equation of the tangent line.
• May 10th 2011, 09:19 PM
RezMan
right but to get y don't I have to plug 1 into the original function. Where then I end up with that wierd number. And then I plug 1 and that number into the dy/dx that I got. And then that other weird number is my slope.
• May 10th 2011, 10:04 PM
RezMan
thanks so much guys! I'm not sure if I'm on the right track so I emailed my teacher. When she gets back to me I'll get back to this.
• May 13th 2011, 06:33 PM
RezMan
hey guys I was able to solve most of it except I'm stuck again. It asks what is the tangent line to the graph at y=0. Give its equation.

tangent line at y=0 is
-4x/9(0)= undefined, right?