Slopes of Parametrized Curves problem

[shivers20]

Find an equation for the line tangent to the curve at the point defined by the given value t. Also find the value of the d^2y/dx^2 at this point.

x= 2 cos t, y= 2 sin t, t= pi/4

This is what I have so far but its wrong.

(2 cos t)/ (-2 sin t)

Set t equal to pi/4

(2 cos45)/(-2 sin45) = sqrt [2]/sqrt [-2] = -1

y+ sqrt[2] = -1 (x- sqrt [2])

the books answer is y = -x + 2sqrt[2]

Where did I go wrong? I apologize for not posting the correct symbols, I wasn;t sure how to. Thanks in advance.

Now for part 2:

Express y prime in terms of t: 2cos(t )/-2sin(t)

Now differentiate with respect to t: 2cos(t )/-2sin(t) nothing changes as of yet

Divide: (2cos(t ))/(-2sin(t)) x 1/(-2sin(t))= ??

 

[galactus]

Find an equation for the line tangent to the curve at the point defined by the given value t. Also find the value of the d^2y/dx^2 at this point.

\(\displaystyle \L\\y''=-2sin(t)\;\ and\;\ x''=-2cos(t)\)

\(\displaystyle \L\\\frac{-2sin(t)}{-2cos(t)}=tan(t)\)

x= 2 cos t, y= 2 sin t, t= pi/4

This is what I have so far but its wrong.

(2 cos t)/ (-2 sin t)

Set t equal to pi/4

(2 cos45)/(-2 sin45) = sqrt [2]/sqrt [-2] = -1

y- sqrt[2] = -1 (x- sqrt [2])

A minor sign reversal

the books answer is y = -x + 2sqrt[2]

Where did I go wrong? I apologize for not posting the correct symbols, I wasn;t sure how to. Thanks in advance.

Now for part 2:

Express y prime in terms of t: 2cos(t )/-2sin(t)

Now differentiate with respect to t: 2cos(t )/-2sin(t) nothing changes as of yet

Divide: (2cos(t ))/(-2sin(t)) x 1/(-2sin(t))= ??