Finding Parametric Equations and Tangent Lines

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Homework Help Overview

The discussion revolves around finding parametric equations and tangent lines for a curve defined by the equations x=1+1/(t^2) and y=1-(3/t) at a specific value of t. Participants are examining the correctness of a tangent line equation and the second derivative.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to find the tangent line and second derivative using the dy/dx formula and point-slope form. Some participants provide calculations for the slope and coordinates at t=2, while others question the correctness of the derivative calculations.

Discussion Status

The discussion includes various calculations and attempts to verify the original poster's results. Some guidance has been offered regarding the method, and a mistake in the derivative calculation has been pointed out, indicating an active exploration of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can share or the methods they can use. There is an emphasis on ensuring the correctness of calculations without providing complete solutions.

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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