Math Is Fun Forum

mathland

Member

Registered: 2021-03-25

Posts: 444

Tangent Line to a Cardioid

The graph of (x^2 + y^2 + y)^2 = (x^2 + y^2)
is a cardioid.

(a) Find all the points on the cardioid that have a horizontal
tangent line. Ignore the origin.

(b) Find all the points on the cardioid that have a vertical tangent
line. Ignore the origin.

Seeking steps for both parts. I must take the derivative on both sides implicitly.
What follows next? Why does the question say to ignore the origin for both parts?

Bob

Administrator

Registered: 2010-06-20

Posts: 10,640

Re: Tangent Line to a Cardioid

It's useful to know what the graph is like.  Here's a link https://www.mathsisfun.com/data/grapher-equation.html

You can copy and paste the equation as it is.

In case you cannot do this I'll also describe it.  It's a (valentine) heart shape coming to a sharp point at (0,0) but more rounded at the bottom.

Differentiation is only allowed where the limit as you approach a point from the left is the same as from the right.  A function for which this is always true has a curvature that changes smoothly.  At a point with a sharp change of direction the gradient is undefined at the point since the left limit and the right limit are different.  That's why the origin is excluded.

Once you have done the implicit differentiation you can substitute dy/dx = 0 to find the horizontal lines.  It looks like there are three points with two sharing the same tangent.

If you make dy/dx the subject of the equation then you can consider what points will make the gradient 'shoot off to infinity'.  That will give the vertical tangents.

Bob

---

Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

mathland

Member

Registered: 2021-03-25

Posts: 444

Re: Tangent Line to a Cardioid

It's useful to know what the graph is like.  Here's a link https://www.mathsisfun.com/data/grapher-equation.html

You can copy and paste the equation as it is.

In case you cannot do this I'll also describe it.  It's a (valentine) heart shape coming to a sharp point at (0,0) but more rounded at the bottom.

Differentiation is only allowed where the limit as you approach a point from the left is the same as from the right.  A function for which this is always true has a curvature that changes smoothly.  At a point with a sharp change of direction the gradient is undefined at the point since the left limit and the right limit are different.  That's why the origin is excluded.

Once you have done the implicit differentiation you can substitute dy/dx = 0 to find the horizontal lines.  It looks like there are three points with two sharing the same tangent.

If you make dy/dx the subject of the equation then you can consider what points will make the gradient 'shoot off to infinity'.  That will give the vertical tangents.

Bob

I'll work on it.