Finding the Tangent Line to a Parametric Curve

[Hobold]

Homework Statement

This is a very basic problem, though it did confuse me a little:

Find the tangent equations to the curve $$x=3t^2+1 \ , \ y = 2t^3+2$$ which intercepts the point (4,3).

Homework Equations

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The Attempt at a Solution

I took $$\frac{dy}{dx} = t = \frac{y-y_0}{x-x_0} = \frac{y-3}{x-4} \rightarrow y=t(x-4)+3$$

What to do next? I don't think the equation works for all t.

 

[Mark44]

Homework Statement

This is a very basic problem, though it did confuse me a little:

Find the tangent equations to the curve $$x=3t^2+1 \ , \ y = 2t^3+2$$ which intercepts the point (4,3).

Are you sure the above is right? If x = 4, then t = +/-1, but when t = 1, y = 4 and when t = -1, y = 0.

Homework Equations

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The Attempt at a Solution

I took $$\frac{dy}{dx} = t = \frac{y-y_0}{x-x_0} = \frac{y-3}{x-4} \rightarrow y=t(x-4)+3$$

What to do next? I don't think the equation works for all t.

 

[Hobold]

Yes, I'm sure, those are the functions.

I believe the problem asks for a tangent of the graphic which will intercept the point (4,3) in R^2, which is not necessarily in the graphic of the function.

 

[Mark44]

OK, I misunderstood.

So let's say we're talking about the point on the curve whose coordinates are (x₀, y₀), that correspond to t = t₀.

Can you use the parametric equations to write x₀ and y₀ in terms of t₀. Then use the point (4, 3) and calculate the slope of the line segment between (x₀, y₀) and (4, 3), which you know is equal to t₀.