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)?

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- Oct 2nd 2011, 05:21 PMCyanBCImplicit 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)? - Oct 2nd 2011, 05:38 PMQuackyRe: Implicit differentiation
Disagree with your differentiation, firstly.

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

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

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

Anyway, you know that $\displaystyle y(1)=7.88811$ so substitute that into the original equation to find the value for x at this point. Then it becomes a case of substitution. - Oct 2nd 2011, 05:50 PMCyanBCRe: 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? - Oct 2nd 2011, 06:01 PMskeeterRe: Implicit differentiation
$\displaystyle \frac{x^2}{36} + \frac{y^2}{64} = 1$

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

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

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

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

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. - Oct 2nd 2011, 06:37 PMCyanBCRe: 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!