The curve C has the equations:
a) find an equation of the tangent at the point where t=2.
What's the procedure for a question like this? Finding dx/dt and dy/dt then subbing in t=2 and then rearranging for an equation in the form y=...x...?
how would I start this thanks
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C4 intergration watch
- Thread Starter
- 11-04-2011 15:34
- 11-04-2011 17:28
- Find dy/dx by the chain rule as you said and sub in t = 2 to get the gradient at that point
- Sub in t = 2 into the parametric equations to find coordinate values (x,y) at the tangent
- Use the formula from c1 , where (x_1, y_1) is your coordinates at the tangent, m is your gradient at the tangent
- 11-04-2011 17:39
Someone asked me to change the title of this thread to "C4 Integration", but since it doesn't actually have anything to do with integration I think I'll let it be.