dy/dx using cosines

[kggirl]

Is this correct?

Find dy/dx if x =cos[t] and y = square of 3 cos[t] and t= 2pi/3

x= -sin2pi/3=
-sin2pi

and the square root of 3 cos 2pi/3 =
square of 3 - sin 2pi/3 = square of 3-sin 2pi

so dy/dx = square root of 3- sin2pi/-sin2pi

the -sin2pi would cancel out, leaving the answer to be the square root of 3

 

[soroban]

Hello, kggirl!

You <u>could</u> be more careful with your notation!

And no, you do <u>not</u> substitute constants before differentiating.
. . If you do, you're differentiating constants, and getting 0 every time.

Find dy/dx if x = cos[t] and y = sqrt{3} cos[t], at t = 2pi/3

First, we find the derivative. . . . . . . . ._
. . . . . . . . . . . . . .dy . . . . dy/dt . . . . -√3 sin t . . . . . _
We know that: . ---- . = . ------- . = . ----------- . = . √3
. . . . . . . . . . . . . .dx . . . . dx/dt . . . . . - sin t
. . . . . . . . . . . . . . . . . _
The slope is <u>always</u> √3 . . . at every value of t.

~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

Knowing this, we can look back and understand why . . .
. . . . . . . . . . . . . . . . . . . . . . . . . . . ._
We have: . (1) x = cos(t), .(2) y = √3 cos(t)
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . _
Substituting (1) into (2), we have: .y .= .√3 x
. . . . . . . . . . . . . . . . . . . . . . . . ._
It is a straight line with slope √3.
. . Of course its derivative is a constant!