please help parameteic equation maths
- Thread starter deviany
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yoscar04
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For 8(iii) you need to use the definition of v: haven't you learnt that v=(dx/dt,dy/dt)?
For 9(ii) how did you get t=2? If you substitute t=2 in your equation dy/dx you never will get 0.
Try to find the t that corresponds to x=1, y=3.
For 9(ii) how did you get t=2? If you substitute t=2 in your equation dy/dx you never will get 0.
Try to find the t that corresponds to x=1, y=3.
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in 9ii, the question asks you to show that at (1,3) the tangent to the curve is parallel to the x-axis, that is, show the slope is 0 at (1,3).
(evaluate something to get 0 and you are done)
what you tried to do is part iii. (solve equation set equal to 0)
(evaluate something to get 0 and you are done)
what you tried to do is part iii. (solve equation set equal to 0)
HallsofIvy
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My neck hurts!
HallsofIvy
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Thank you! That helps a lot.
Problem 9 gives parametric equations \(\displaystyle x= e^{3t}\), \(\displaystyle y= t^2e^t+ 3\). \(\displaystyle dx/dt= 3e^{3t}\), \(\displaystyle dy/dt= 2te^t+ t^2e^t= e^t(t^2+ 2t)\). \(\displaystyle \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{e^t(t^2+ 2t)}{3e^{3t}}= (t^2+ 2t)e^{-2t}\). That is the same as the given \(\displaystyle \frac{t(t+ 2)}{3e^{2t}}\).
At x= 1, y= 3, \(\displaystyle x= 1= e^{3t}\) so t= 0 (as a check \(\displaystyle y= 0^2e^0+ 3= 3\)). With t= 0, the derivative, whether we use \(\displaystyle (t^2+ 2t)e^{-2t}\) or \(\displaystyle \frac{t(t+ 2)}{3e^{2t}}\), is 0 so the tangent line is horizontal, parallel to the x- axis.
The tangent line will be horizontal wherever \(\displaystyle dy/dx= \frac{t(t+ 2)}{3e^{2t}}= 0\). A fraction is 0 if and only if the numerator is 0: t(t+ 2)= 0 so either t= 0 or t= -2. The other point where the tangent line is parallel to the x-axis is \(\displaystyle \left(e^{3(-2)}, (-2)^2e^{-2}+ 3\right)= (e^{-6}, 4e^{-2}+ 3)\).
Problem 9 gives parametric equations \(\displaystyle x= e^{3t}\), \(\displaystyle y= t^2e^t+ 3\). \(\displaystyle dx/dt= 3e^{3t}\), \(\displaystyle dy/dt= 2te^t+ t^2e^t= e^t(t^2+ 2t)\). \(\displaystyle \frac{dy}{dx}= \frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{e^t(t^2+ 2t)}{3e^{3t}}= (t^2+ 2t)e^{-2t}\). That is the same as the given \(\displaystyle \frac{t(t+ 2)}{3e^{2t}}\).
At x= 1, y= 3, \(\displaystyle x= 1= e^{3t}\) so t= 0 (as a check \(\displaystyle y= 0^2e^0+ 3= 3\)). With t= 0, the derivative, whether we use \(\displaystyle (t^2+ 2t)e^{-2t}\) or \(\displaystyle \frac{t(t+ 2)}{3e^{2t}}\), is 0 so the tangent line is horizontal, parallel to the x- axis.
The tangent line will be horizontal wherever \(\displaystyle dy/dx= \frac{t(t+ 2)}{3e^{2t}}= 0\). A fraction is 0 if and only if the numerator is 0: t(t+ 2)= 0 so either t= 0 or t= -2. The other point where the tangent line is parallel to the x-axis is \(\displaystyle \left(e^{3(-2)}, (-2)^2e^{-2}+ 3\right)= (e^{-6}, 4e^{-2}+ 3)\).