Finding Parametric Equations and Tangent Lines

AI Thread Summary
The discussion focuses on finding the parametric equations and tangent lines for the curve defined by x=1+1/(t^2) and y=1-(3/t) at t=2. The proposed tangent line equation is y=-6x+7, with the second derivative calculated as d^2y/dx^2=24. The method used involves the dy/dx formula and point-slope form, but a correction is noted regarding the derivative dx/dt, which should be -2/t^3. Overall, the calculations are mostly correct, but attention to detail in derivatives is crucial for accuracy.
MozAngeles
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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
 
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Your method is correct. Show your calculation in detail.

ehild
 
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
 
You have a mistake: dx/dt=-2/t3.

ehild
 
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