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Consider the curve defined by the parametric equations $x=t^2 +t-1$ and $y=te^{2t}$

i) Show that $dy/dx =e^{2t}$

ii) Hence show that the tangent to the curve at the point on the curve where $t= -1$ passes through the origin.

I'm sorry to bug you guys, but I'm clueless and would help me if someone could help me, so I get questions that are similar to this. Thanks!

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for the first part, use the fact that dy/dx = (dy/dt)/(dx/dt). –  symplectomorphic Jul 8 '13 at 6:40
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1 Answer

up vote 3 down vote accepted

$$\frac{dy}{dt}=e^{2t}+2te^{2t}=e^{2t}(1+2t)$$

$$\frac{dx}{dt}=2t+1 $$

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

At $t=-1,x=-1, y=-e^{-2}$

So, the equation of the tangent will be $$\frac{y-(-e^{-2})}{x-(-1)}=\frac{dy}{dx}_{\text{(at }t=-1)}=e^{-2}\implies y=x\cdot e^{-2}$$ which clearly passes through the origin $(0,0)$

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THANKS, your a genius! –  Red Queen10101 Jul 8 '13 at 7:19
    
@RogerShan, my pleasure. –  lab bhattacharjee Jul 8 '13 at 7:40
    
Excuse me, I would like to ask how do you mark a question to be answered? sorry Im new with this site. –  Red Queen10101 Jul 8 '13 at 9:17
    
hahaha dw got it –  Red Queen10101 Jul 8 '13 at 9:17
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