Hi all, first time poster but if this goes well will probably be seeking out more help.
So we are doing parametric equations and I understand how to find the slope, however from their I'm totally lost. >.<
Please don't just answer, I have an exam soon and if you could explain why, it would be wonderful.
The original question is:
Find an equation of the tangent line to this curve
x=(t^2)-8t
y=(t^2)+7t
So I know you take the derivative of both sides to find dx/dt and dy/dt
dx/dt=2t-8
dy/dt=2t+7
So the slope is equal to...
(2t+7)/(2t-8)
The T value they give is 8/2 (4?)
So..
2(4)+7/2(4)-8
15/0
Slope is undefined....
From there I have no idea
_______________________
However if the slope were to work, such as with this equation...
x=(t^3)-3t
y=t^2
@ t= 3
I get the slope 1/4
so
y=(x/4)+b
However, I don't know what values to put for x and y!!
Anyway. Thank you for any help.