# Parametric Curves

• Apr 25th 2007, 10:21 AM
Xavier20
Parametric Curves
"Given the Parametric equations x=tsint and y=tcost find dy/dx at the point ((pi/2),0)

I remember that one must differentiate y'/x', but I don't remember what to do with the given point.

Y' = cost + -tsint
X' = sint + tcost

Do you plug in pi/2 into X', and 0 into Y', or?
• Apr 25th 2007, 11:28 AM
CaptainBlack
Quote:

Originally Posted by Xavier20
"Given the Parametric equations x=tsint and y=tcost find dy/dx at the point ((pi/2),0)

I remember that one must differentiate y'/x', but I don't remember what to do with the given point.

Y' = cost + -tsint
X' = sint + tcost

Do you plug in pi/2 into X', and 0 into Y', or?

dy/dx = dy/dt dt/dx

Now differentiate x=t sin(t) wrt x to get:

1 = dt/dx sin(t) + t cos(t) dt/dx

so:

dt/dx = 1/[sin(t) +y]

Now differentiate y= t cos(t) wrt t to get:

dy/dt = cos(t) - t sin(t) = cos(t) - x

so dy/dx = [cos(t) - x]/[sin(t) +y]

Now the point ((pi/2),0) corresponds to t=p/2 and plugging this into the
above gives us dy/dx at this point is:

[-pi/2]/[1] =-pi/2

RonL
• Apr 25th 2007, 12:37 PM
Soroban
Hello, Xavier!

You have the right idea . . . but be careful.

Quote:

Given the parametric equations: .x = t新in(t) and y = t搾os(t)
find dy/dx at the point (π/2 ,0)

. . . . . . . . . . . . . .dy . . . dy/dt
The derivative is: . --- .= .-------
. . . . . . . . . . . . . .dx . . . dx/dt

Your derivatives were correct: .dy/dt .= .cos(t) - t新in(t)
. . . . . . . . . . . . . . . . . . . . . .dx/dt .= .sin(t) + t搾os(t)

. . . .dy . . . cos(t) - t新in(t)
So: .--- .= .-------------------
. . . .dx . . . sin(t) + t搾os(t)

The point (π/2, 0) means: .x = π/2, y = 0

. . Hence: .t新in(t) = π/2 . . t = π/2

. . . . . . . . .dy . - . cos(π/2) - (π/2)新in(π/2) . - . .0 - (π/2)(1) . . . . .π
Therefore: . --- .= .------------------------------ .= .--------------- .= .- --
. . . . . . . . .dx . . . sin(π/2) + (π/2)搾os(π/2) . . . 1 + (π/2)(0) . - . . 2

• Apr 25th 2007, 03:35 PM
Xavier20
Thank you! =)