Question

HELP PLEASE

Answer 1

I need help doing #6.

Answer 2

@zepdrix @518nad

Answer 3

hmm i think you would find the derivative and set it = to 0
find the derivative of x= t^3-3t

Answer 4

This one is not as straightforward as it might appears on first glad. It involves deriving but you need to know what to do after that.

Here is the general gist - when you parameterise an equation, you tend to generate two functions.

y = g(t) and x = h(t) for some real parameter t.
However dy/dt and dx/dt give you only snapshots of how these are changing based on the parameter and not the original (Cartesian function). You need to convert to Cartesian form generally to understand what is happening between y and x.

However it is very hard to change the parameterised form to get a Cartesian equation.
It is however easy to differentiate each individually, then apply the chain rule...

y = t^2 - 9 --> dy/dt = 2t
x= t^3 - 3t --> dx/ dt = 3(t^2 - 1)

The chain rule:
\[\frac{ dy }{ dx } = \frac{\frac{ dy }{ dt } }{ \frac{ dx }{ dt } } = \frac{ 2t }{ 3(t^2-1) }\]

Now we can understand what is happening better. dy/dx = 0 means we have a stationary pt or horizontal tangent somewhere while dy/dx being undefined (when denominator tends to zero) yields a vertical tangent.

So solve the numerator for zero to find the value of our parameter t, when we attain horizontal tangent and sub into y(t) and x(t). Do the same with the denominator.

In fact your plot looks like this and should match your points algebraically. https://www.wolframalpha.com/input/?i=parametric+plot+(t%5E3-3*t,+t%5E2+-+9)

Answer 5

What do you mean by solve the numerator by 0?
Are you saying to set dy = 0?

Answer 6

By chain rule dy/dx = (dy/dt) / (dx/dt) and dy/dt = 2t, dx/dt = 3(t^2-1)

So the numerator would be dy/dt = 2t and denominator would be dx/dt = 3(t^2-1)

Answer 7

Yes i understand that part.

Answer 8

Now use the formula
\[\frac{dy}{dx}=\frac{dy}{dt}\times\frac{dt}{dx}\]
to find dy/dx

Answer 9

To then find horizontal tangent, set dy/dx to 0. Work out the t value and then work out x and y coordinates
To find vertical, set the denominator of dy/dx to 0. Work out the t value and then work out x and y coordinates

Answer 10

so you have
\[\frac{ dy }{ dx } = \frac{ 2t }{3( t^2 -1) }\]
\[\frac{ dy }{ dx } = 0 ~(stationary ~~pt) \rightarrow 2t = 0, ~t = 0\]
Sub that t back into y(t) and x(t) parametric forms

Then do the same but make the denominator which is 3t^2-3 = 0 so that dy/dx tends to infinity on either side.

Answer 11

Sid, he's saying,\[\large\rm \frac{dy}{dx}=0\qquad\implies\qquad 2t=0\]

and\[\large\rm \frac{dy}{dx}=undefined\qquad\implies\qquad 3(t^2-1)=0\]

You're separately setting the numerator and denominator equal to zero to find different types of critical points.

Answer 12

OOOOOOH.

Answer 13

So then would t=-1,0,1 ?

Answer 14

The t values will be separate for both cases.
For horizontal set t = 0
For vertical set t = -1 and t =1

Answer 15

Oh. Okay, and then how do I find the (x,y) coordinates?

Answer 16

You are given x and y equations in terms of t.
Just input the t values in those equations.

Answer 17

Oh. okay

Answer 18

yeah sorry if I wasn't clear enough, the t values come off dy/dx and you set that back into the original parametric forms. Thanks @zepdrix

Answer 19

Apologies @mww and @zepdrix for interrupting in your explanations.

Answer 20

nah it's cool. everyone contributed to his understanding

Answer 21

Okay.... So....
I got (0,-9) for the Horizontal Tangent and got that right.

But when I plugged in 1 & -1 back into the equation, I got 2 & -2 for my X values and -8 for my Y value.

I tried (-2,8) and (2,8)... But got both points for the vertical tangent wrong.

Answer 22

try (-2, -8)and

(2,-8 )

Answer 23

y = t^2 - 9 = (1)^2 - 9 = -8...

Answer 24

Sorry I meant -8.

Answer 25

I plugged in -8 not 8

Answer 26

(-2,-8) and (2,-8) did not work

Answer 27

there is a flaw with the site. I checked on a graphing calculator. The answer is in fact (-2,-8)