please help parameteic equation maths

[deviany]

number 9 part 2 and 3. if its parallel to x axis then the gradient would be 0. but how can i proof it? i triedy dy/dx=0 but i got t=2... which part i am wrong?

 

[yoscar04]

Do you know how to define velocity, given its position (x,y)?

 

[yoscar04]

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.

 

[Singleton]

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)

 

[LCKurtz]

Please post images right side up.

 

[HallsofIvy]

Please post images right side up.

My neck hurts!

 

[yoscar04]

For the benefit of the public:

 

[HallsofIvy]

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 $\displaystyle \left(e^{3(-2)}, (-2)^2e^{-2}+ 3\right)= (e^{-6}, 4e^{-2}+ 3)$.