Derivative of a curve and tangent lines.

2009-11-28T17:33:51Z

I'm having trouble with the last question on my AP calc homework. Can anyone show me how it's done?

Consider the curve given by $x^2 + 3y^2 = 1 + 3xy$

a) show that $\frac{dy}{dx} = \frac{3y - 2x}{6y - 3x}$

b) find all points on the curve whose x-coordinate is 1, and write an equation for the tangent line at each of these points.

c) find the coordinates of each point where the tangent line is vertical

Answer by notsobright for Derivative of a curve and tangent lines.

2009-11-29T16:22:02Z

I got it.

First the differentiation:

$\frac{d}{dx}x^2+3 y(x)^2 = \frac{d}{dx}3xy(x)+1$

Differentiate the sum term by term and factor out constants:

$\frac{d}{dx}x^2+3 \frac{d}{dx}y(x)^2 = \frac{d}{dx}3xy(x)+1$

The derivative of $x^2$ is $2x$:

$3\frac{d}{dx}y(x)^2+2x = \frac{d}{dx}3xy(x)+1$

Use the chain rule:

$\frac{d}{dx}y(x)^2 = \frac{du^2}{du} \frac{du}{dx}$
where $u = y(x)$ and $\frac{du^2}{du} = 2 u$:

$6y(x) \frac{d}{dx}y(x)+2x = \frac{d}{dx}3xy(x)+1$

The derivative of $y(x)$ is $y'(x)$:

$6y(x)y'(x)+2x = \frac{d}{dx}3xy(x)+1$

Differentiate the sum term by term and factor out constants:

$6y(x) y'(x)+2x = 3 \frac{d}{dx}xy(x)+\frac{d}{dx(1)}$

The derivative of 1 is zero:

$6y(x) y'(x)+2x = 3 \frac{d}{dx}xy(x)$

Use the product rule,
$\frac{d}{dx(u v)} = v \frac{du}{dx}+u\frac{dv}{dx}$
where $u = x$ and $v = y(x)$:

$6 y(x)y'(x)+2x = 3 (x \frac{d}{dx}y(x)+y(x) \frac{d}{dx(x)})$

The derivative of $y(x)$ is $y'(x)$:

$6 y(x)y'(x)+2x = 3y(x) \frac{d}{dx(x)}+3xy'(x)$

The derivative of $x$ is 1:

$6 y(x)y'(x)+2x = 3xy'(x)+3y(x)$

To find where $x=1$ just put $x=1$ into the original equation to get the y equation $y^2+y=0$ where the roots are $y=1$ or $y=0$

The tangent is vertical where $\frac{dy}{dx}=$undefined (or infinite)