Finding Parametric Equations and Tangent Lines

[MozAngeles]

Homework Statement

So I'm studying for my test. doing even and odd problems from the book. I wanted to see if this answer is right.
Q: find an equation for the line in the xy-plane that is tangent to the curve at the point corresponding to the given value of t.Also, find the second derivative
x=1+1/(t^2), y=1-(3/t); t=2

A:line y=-6x+7
d^2y/dx^2=24
So if its not right can you point me in the right direction.

Homework Equations

The Attempt at a Solution

i used the dy/dx formula which is (dy/dt)/(dx/dt), then put it in point slope form. then used the formula for the second derivative

 

[ehild]

Your method is correct. Show your calculation in detail.

ehild

 

[MozAngeles]

slope=dy/dx=3/t²÷ -1/t³= -3t, plugging in t=2, i got slope=-6

then plugging t=2 into x=1+1/t² i got 5/4
and into y=1-3/t i got -1/2

plugg those into y-y₁=slope(x-x₁), then simplyfying i got -6x+7

then for d²y/dx²=dy^'/dy÷dx/dt
y^'=-3t
dy^'=-3
so plugging back into
dy^'/dy÷dx/dt=-3/(-1/t³)=3t³= 24 after plugging t=2

 

[ehild]

You have a mistake: dx/dt=-2/t³.

ehild

 

[rwisz]

which is d^2y/dx^2= (d/dt)(dy/dx).

Good job on using the correct formulas for finding the tangent line and second derivative. Your equation for the tangent line, y = -6x + 7, is correct. To check your work, you can plug in the given value of t (t=2) into the original parametric equations and see if the point (1, -1) lies on the line. As for the second derivative, you correctly found that d^2y/dx^2 = 24. Keep up the good work in studying for your test!