graphs - how do you know when a cusp forms? (1 Viewer)

[ninetypercent]

so for the curve y^2 = x^3, a cusp forms at x = 0

but how do you know when this forms?

Thnx

 

[Carrotsticks]

When dy/dx is undeterminable. ie: 0/0.

2yy'=3x^2

y' = 3x^2/2y

At origin (0,0), y' = 0/0, which is undeterminable. Therefore, a cusp.

 

[jet]

When dy/dx is undeterminable. ie: 0/0.

2yy'=3x^2

y' = 3x^2/2y

At origin (0,0), y' = 0/0, which is undeterminable. Therefore, a cusp.

That doesn't necessarily imply a cusp, as it could just as easily be a vertical tangent without you knowing.

If we were to express y in terms of x we would have
We can stick this into the derivative to get

Thus you'll have two different derivatives for each value of x. As x approaches 0, \frac{dy}{dx} approaches 0. Because you have the matter of the derivative taking different signs, you get this type of function where the derivative has a different value on either side of the point and it doesn't exist. (Note that this is a bit higher than the scope of the 4 unit course).

Ideally you want opposite signed derivatives on either side of the point/two derivatives at the point to give you a cusp.

Take for example y = |x|. We don't know how to differentiate this but we can define it as a piecewise function:


Then we can differentiate it on both sides of x = 0 and we get two different derivatives: y = 1 for x ≥ 0 and y = -1 for x < 0. Obviously we have a derivative with two different values on either side of x = 0 and so there is a cusp there.