Implicit Differentiation, Tangent Lines

**kinana18** · January 25th 2009, 02:46 PM

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).

**chiph588@** · January 25th 2009, 02:58 PM

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

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

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

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

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

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

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

**chiph588@** · January 25th 2009, 03:04 PM

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

We have $m$ from part a.

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

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

**chiph588@** · January 25th 2009, 03:12 PM

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

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

$3y+3 = x+4y$

$y = 3-x$

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

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

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