Finding Parametric Equations and Tangent Lines

  • Thread starter MozAngeles
  • Start date
  • #1
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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
 

Answers and Replies

  • #2
ehild
Homework Helper
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Your method is correct. Show your calculation in detail.

ehild
 
  • #3
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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
 
  • #4
ehild
Homework Helper
15,543
1,914
You have a mistake: dx/dt=-2/t3.

ehild
 

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