Parametric equations for a particle

[rjs123]

Homework Statement

The path of a particle is given for time t > 0 by the parametric equations: x = t +2/t, y = 3t^2

a. Find the coordinates of each point on the path where the velocity of the particle in the x direction is zero.

b. Find dy/dx when t = 1

c. Find d^2y/dx^2 when y = 12

The Attempt at a Solution

a. No horizontal tangents...dy/dt = 6t...= 0 when t = 0...but t must be greater than 0 (condition)

b. dy/dx = (dy/dt)/(dx/dt) = 6t/(1-2/t^2) = -6...when you plug in 1

c. stuck on c at the moment

 

[rock.freak667]

For a) you want to get were dx/dt = 0 and then rearrange for t. This will give a value for t, so you can get the x and y coordinates for it.

b) is correct.

For part c, differentiate the expression wrt t using implicit differentiation, can you do this?

 

[rjs123]

a. dx/dt = 1 - 2/t^2 = 0

1 = 2/t^2

1/2 = t^2

t = sqrt(1/2)


x = 3.5355, y = 1.5


this is what i got for your recommendation.

 

[rock.freak667]

a. dx/dt = 1 - 2/t^2 = 0

1 = 2/t^2

1/2 = t^2

t = sqrt(1/2)


x = 3.5355, y = 1.5


this is what i got for your recommendation.

I thought when the velocity of the particle in the x direction is zero when there is a horizontal tangent present?

The gradient function of x is dx/dt. So if dx/dt = 0 (or the velocity in the x-direction is zero), then the tangent drawn has a zero gradient i.e. is a horizontal tangent.

 

[rjs123]

The gradient function of x is dx/dt. So if dx/dt = 0 (or the velocity in the x-direction is zero), then the tangent drawn has a zero gradient i.e. is a horizontal tangent.

yes, the graph has no points where there could be a horizontal tangent line...the one found on your recommendation contains a slope of -2...at (3.5, 1.5)

 

[rock.freak667]

yes, the graph has no points where there could be a horizontal tangent line...the one found on your recommendation contains a slope of -2...at (3.5, 1.5)

if x = t + 2/t had no stationary points then putting dx/dt = 0 would lead to something like 0=2.

However there would be a slope since the vertical velocity is not zero at t = 1.

 

[HallsofIvy]

Homework Statement

The path of a particle is given for time t > 0 by the parametric equations: x = t +2/t, y = 3t^2

Is x= t+ (2/t) or (t+2)/t?

a. Find the coordinates of each point on the path where the velocity of the particle in the x direction is zero.

b. Find dy/dx when t = 1

c. Find d^2y/dx^2 when y = 12

The Attempt at a Solution

a. No horizontal tangents...dy/dt = 6t...= 0 when t = 0...but t must be greater than 0 (condition)

b. dy/dx = (dy/dt)/(dx/dt) = 6t/(1-2/t^2) = -6...when you plug in 1

c. stuck on c at the moment

 

[SammyS]

For (a): If the velocity in the x direction is zero, then the tangent to the path is vertical, not horizontal.

For (c):

\frac{d^2y}{dx^2}=\frac{d\left(dy/dx\right)}{dx}=\frac{d\left(dy/dx\right)/dt}{dx/dt}​

and you found that

\frac{dy}{dx}=\frac{6t}{1-2t^{-2}}​