Question

How to determine dy / dx, d^2y/dx^2, for which curve values is concave upwards?:

`x=t^2/2+t, y=t^2/2-t`

Answer 1

This reference, Tangents with Parametric Equations, tells us how to compute `dy/dx` and `(d^2y)/dx^2`

Explanation:

Compute `dx/dt`:

`dx/dt = t+1`

Compute `dy/dt`

`dy/dt = t-1`

Use the following equation taken from the reference:

`dy/dx = (dy/dt)/(dx/dt)`

Substitute our computations:

`dy/dx = (t-1)/(t+1)`

Use the following equation taken from the reference:

`(d^2y)/dx^2= ((d(dy/dx))/dt)/(dx/dt)`

To compute `(d(dy/dx))/dt`, we use the quotient rule:

`(d((t-1)/(t+1)))/dt = ((d(t-1))/dt(t+1)- (d(t+1))/dt(t-1))/(t+1)^2`

`(d((t-1)/(t+1)))/dt = (1(t+1)- 1(t-1))/(t+1)^2`

`(d((t-1)/(t+1)))/dt = (t+1-t+1)/(t+1)^2`

`(d((t-1)/(t+1)))/dt = 2/(t+1)^2`

`(d^2y)/dx^2= (2/(t+1)^2)/(t+1)`

`(d^2y)/dx^2= 2/(t+1)^3`

This reference, The Shape of a Graph, Part 2, tells us that the graph is concave up when the second derivative is positive:

`2/(t+1)^3>0 larr` This is true for `t > -1`