# Tangent Lines of Parametric Equations

1. Oct 31, 2007

### Feldoh

My book really doesn't go into a lot of depth but I was wondering if this is correct

If we are asked to find the tangent line of a specific value of t for a given parametric equation then we can find the equation of the tangent line in either rectangular or parametric functions.

Rectangular Mode
We need dy/dx and the point at the specific t value, say $(x_o,y_o)$ is our point.
The tangent line is:

$$y-y_o = \frac{dy}{dx}(x-x_o)$$

Parametric Mode
We need dy/dt, dx/dt, and the point at the specific t value, once again say $(x_o,y_o)$.

$$x(t) = \frac{dx}{dt}t+x_o$$

$$y(t) = \frac{dy}{dt}t+y_o$$

Is that correct?

Also could someone explain how we derive dy/dx and d^2y/dx^2?

2. Oct 31, 2007

### EnumaElish

Yes.
From the slope of the function to which the tangent line is tangent at (x_0, y_0).

3. Nov 1, 2007

### Feldoh

Ok, thanks^^

I should be more specific:

My text gives this definition for dy/dx

"The Chain Rule states that the derivative dy/dx for the parametric curve is the ratio of dy/dt to dx/dt." But I cannot see where they got that from. They then go on to say "d^2y/dx^2 can be derived in the same manner". Could you explain it a little more, I'm a bit confused? >.>

4. Nov 2, 2007

### EnumaElish

Let D be the partial derivative operator.

If y(t) = f(x(t)), or with some abuse of notation y(t) = y(x(t)), how do you write Dy/Dt?

5. Nov 3, 2007

### HallsofIvy

The chain rule. dy/dx= dy/dt dt/dx= (dy/dt)/(dx/dt). Saying "d^2y/dx^2 can be derived n the same manner" is little misleading- it's much more complicated. Notice tha they don't give that formula!

d^2y/dx^2= d(dy/dx)/dx= (1/(dx/dt)) d((dy/dt)/(dx/dt))/dt and you have to use the quotient rule for the last part.

6. Nov 4, 2007

### Feldoh

Ah seeing y(t) = f(x(t)) makes it clear to me

$$\frac{dy}{dt} = \frac{dy}{dx}*\frac{dx}{dt}$$

Which we rearrange to get:

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}$$