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)?
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.
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?
$\displaystyle \frac{x^2}{36} + \frac{y^2}{64} = 1$
$\displaystyle \frac{y^2}{64} = 1 -\frac{x^2}{36}$
$\displaystyle y^2 = 64\left(1 -\frac{x^2}{36}\right)$
$\displaystyle y = \pm 8\sqrt{1 -\frac{x^2}{36}}$
$\displaystyle y(1) = \pm 8\sqrt{1 -\frac{1}{36}} = \pm 8\sqrt{\frac{35}{36}} \approx \pm 7.88811$
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.