1. ## Implicit differentiation

suppose (x^2/26)+(y^2/64)=1 and y(1)=7.88811 Find y'(1) by implicit diffentiation

I get y'=16x/9y but I'm not sure where to go from there to get y'(1)?

2. ## Re: Implicit differentiation

$\frac{x}{13}+\frac{y}{32}\cdot\frac{dy}{dx}=0$

$\frac{x}{13}=-\frac{y}{32}\cdot\frac{dy}{dx}$

$\frac{dy}{dx}=\frac{-32x}{13y}$

Anyway, you know that $y(1)=7.88811$ so substitute that into the original equation to find the value for x at this point. Then it becomes a case of substitution.

3. ## Re: Implicit differentiation

Sorry I typoed
Should have been
(x^2/36)+(y^2/64)=1

Forgive me, but I'm just not making a connection between the information y(1)=7.88811 and the original equation. I know I need to substitue and solve for x, but there's a point in my brain that doesnt want to accept how this should occur.

is it just plug in 7.88811 for y^2 and solve for x?

4. ## Re: Implicit differentiation

Originally Posted by CyanBC
Sorry I typoed
Should have been
(x^2/36)+(y^2/64)=1

Forgive me, but I'm just not making a connection between the information y(1)=7.88811 and the original equation. I know I need to substitue and solve for x, but there's a point in my brain that doesnt want to accept how this should occur.

is it just plug in 7.88811 for y^2 and solve for x?
$\frac{x^2}{36} + \frac{y^2}{64} = 1$

$\frac{y^2}{64} = 1 -\frac{x^2}{36}$

$y^2 = 64\left(1 -\frac{x^2}{36}\right)$

$y = \pm 8\sqrt{1 -\frac{x^2}{36}}$

$y(1) = \pm 8\sqrt{1 -\frac{1}{36}} = \pm 8\sqrt{\frac{35}{36}} \approx \pm 7.88811$

they chose to use the positive value of y on the ellipse.

so, correctly determine dy/dx and sub in 7.88811 for y and 1 for x to get the slope at that point on the ellipse.

5. ## Re: Implicit differentiation

Thank you both very much. That one was driving me crazy. I just need to relax and think a bit more clearly when I get these.

Solution.
-16x/9y is -16(1)/9(7.88811) or -0.225374364426685

Again, thank you very much!