# Thread: Implicit Differentiation, Tangent Lines

1. ## Implicit Differentiation, Tangent Lines

Given the curve x+xy+2y^2=6.
a)Find an expression for the slope of the curve at any point (x,y) on the curve.
b)Write an expression for the line tangent to the curve at the point (2,1).
c)Find the coordinates of all other points on this curve with slope equal to the slope at (2,1).

2. a.) $\displaystyle x+xy+2y^2 = 6$

$\displaystyle \frac{d}{dx} (x) + \frac{d}{dx} (xy) + \frac{d}{dx} (2y^2) = \frac{d}{dx} (6)$

Note that due to the chain rule $\displaystyle \frac{d}{dx} (f(y)) = f'(y) \frac{dy}{dx}$

$\displaystyle 1 + \left( x\frac{dy}{dx} + y \right) +4y\frac{dy}{dx} = 0$

$\displaystyle x\frac{dy}{dx} + 4y\frac{dy}{dx}=-1-y$

$\displaystyle \frac{dy}{dx}(x+4y) = -(y+1)$

$\displaystyle \frac{dy}{dx} = -\frac{y+1}{x+4y}$

3. b.) $\displaystyle y-y_0 = m(x-x_0)$ where $\displaystyle (x_0,y_0) = (2,1)$

We have $\displaystyle m$ from part a.

$\displaystyle m = -\frac{y_0+1}{x_0+4y_0}$

now just plug in the values for $\displaystyle x_0$ and $\displaystyle y_0$ and solve for $\displaystyle y$ and you're done.

4. c.) from part b we have that $\displaystyle m = -\frac{1}{3}$, so just set $\displaystyle \frac{dy}{dx} = -\frac{1}{3}$

$\displaystyle -\frac{y+1}{x+4y} = -\frac{1}{3}$

$\displaystyle 3y+3 = x+4y$

$\displaystyle y = 3-x$

So the slope of the curves tangent line is the same when [tex] y = 3-x [tex] and $\displaystyle x+xy+2y^2 = 6$.

Solving this system of equations we get $\displaystyle x=6, \; y=-3$