Question

some what simple parametric equations question

find the points on the cruve where the tanget is horizontal or vertical.

x=t^3-3t y=t^2-3

i know how to find the derivative, i get (2t)/(3t^2-3)

the tangent should be vertical when t= negative or postive 1 right? but my book says negative or postive 2 for some reason. and i get that the tanget is horzontal at x=0, since that would make the numerator 0.

Answer 1

sec, i just realized i mistyped

Answer 2

it should be horizontal at the point x=0, and vertical at x= plus or minus 2.

Answer 3

You know the equation for \(dy/dx\) right?
So find out when numerator is 0 (horizontal)
And when denominator is 0 (vertical)

Answer 4

When both are 0 then.. well things get tricky.

Answer 5

denominator =0 when x =negative of postive 1. but in my book it says negative or positive 2.

Answer 6

negative or positive*

Answer 7

did you write the problem correctly?

Answer 8

x=t^3-3t and y=t^2-3

is exactly how its written in the book

Answer 9

maybe i took the dy/dx incorrectly, but i dont see any mistakes. its just a matter of taking the derivative of each parametric equation, then putting the dy over dx. (dy/dx)

Answer 10

maybe the book is wrong haha? doubt it though ._.

Answer 11

Hmmmm, what happens at 2 and -2 anyway?

Answer 12

not entirely sure. but according to the book thats where the tangent line is vertical.

Answer 13

Hmmm, maybe you are answering in terms of \(t\) but should answer in terms of \(x\)

Answer 14

What is \( x(t=-1)\) and \(x(t=1)\)

Answer 15

when t=1, x=-2 and when t=-1, x =2

Answer 16

ic now, guess i had to plug values back in to parametric equations

Answer 17

ty