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# c4 differentiation help watch

1. Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2
I managed to get the answer as y=2 but the answer is apparently x=1?

2. (Original post by doglover123)
Q3a). Find the equation of the normal to the curve with parametric equations x=e^t, y=e^t + e^-t, at point P, where t=2
I managed to get the answer as y=2 but the answer is apparently x=1?

Could you provide some working please so we can identify where you went wrong?

Edit 1: Neither x=1 nor y=2 are going to be right as the normal will be oblique and those lines do not pass through P.

Edit 2: Did you mean t = 0 and did you find the tangent rather than the normal?
3. (Original post by Mr M)
Could you provide some working please so we can identify where you went wrong?

Edit 1: Neither x=1 nor y=2 are going to be right as the normal will be oblique and those lines do not pass through P.

Edit 2: Did you mean t = 0 and did you find the tangent rather than the normal?
I'm sorry, I was looking at the q before, t does equal 0.

dy/dt = e^t-e^-t/e^t
substitute t=0 dy/dx = 0 therefore gradient of normal = 0 and (1,20
y-y1=m(x-x1)
y-2=0
y=2
4. (Original post by doglover123)
I'm sorry, I was looking at the q before, t does equal 0.

dy/dt = e^t-e^-t/e^t
substitute t=0 dy/dx = 0 therefore gradient of normal = 0 and (1,20
y-y1=m(x-x1)
y-2=0
y=2
Right firstly you don't mean dy/dt you mean dy/dx. But dy/dx is the gradient of the tangent not the normal. If the tangent is horizontal (this is what you have just found) then the normal will be vertical.
5. (Original post by Mr M)
Right firstly you don't mean dy/dt you mean dy/dx. But dy/dx is the gradient of the tangent not the normal. If the tangent is horizontal (this is what you have just found) then the normal will be vertical.
Ahhh I didn't know that! Thanks for clearing things up, also I apologise for the typo's, I should have proof read. Thanks again!

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Updated: January 11, 2017
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