# Homework Help: Implicit differentiation and a bicorn curve

1. Mar 7, 2005

### erik05

Hey all. There's a question I seem to be stuck on involving implict differentiation. Here it is:

The curve called a bicorn has the equation $$(x^2+8y-16)^2=y^2(16-x^2)$$ Verify by implicit differentiation that its tangent lines at the points $$(0,4)$$ and $$(0,\frac{4}{3})$$ are horizontal. Do by: a) expanding the equation
b) not expanding the equation

a) I expanded out the equation to get $$x^4+ x^2y^2 + 16x^2y-32x^2+48y^2-256y+256$$ and took the derivative of all the terms and got the answer $$y'= \frac {-4x^3-2xy^2-32xy+64x}{2x^2y+16x^2+96y-256}$$

b) Not expanding the equation and using the chain rule and product rule for each side respectively and then taking the derivative, I also got $$y'= \frac {-4x^3-2xy^2-32xy+64x}{2x^2y+16x^2+96y-256}$$

My question is now,how would you solve the above derivative to get the points $$(0,4)$$ and $$(0,\frac {4}{3} )$$ to verify it? Or would you just put in those two points in the derivative and assuming that I did it right, get the slope of 0? I'm pretty sure that the better way to do this is to solve for those two points since you could just put those points in the original equation $$(x^2+8y-16)^2=y^2(16-x^2)$$ to get zero but correct me if I'm wrong. So,any suggestions? Thank you for the help in advance.

P.S- Not that it really matters but here is a picture of a bicorn:
http://astronomy.swin.edu.au/~pbourke/curves/bicorn/1.gif [Broken]

Last edited by a moderator: May 1, 2017
2. Mar 7, 2005

### dextercioby

You have to compute the slope at those specific points ,which means that u have to eveluate the derivative at those points ...

Daniel.

P.S.BTW,this fact is trivial.

3. Mar 7, 2005

### erik05

So basically what I was doing was wrong?

4. Mar 7, 2005

### cepheid

Staff Emeritus
^ doesn't make any sense. Why do you seem to think that you will somehow "get those points" out of that expression?

Exactly. That is the correct method. THINK about why for a minute. What information have you been given? That the slope of a tangent line to the curve at each of the points (0,4) and (0, 4/3) is zero. That means that the derivative of the function is zero at those points i.e. the function y'(x) has a value of of 0 when x = 0 and y = 4, and also when x = 0 and y = 4/3. Substitute those x and y values into your expression for the derivative in order to compute the value of the derivative at that point.

Sorry, like I said, this makes no sense.

What makes you say that? EDIT: that expression is not even an explicit expression for the function y(x), so plugging those two numbers in will give you nothing useful at all. You'll just get:

something = something (which is correct, but not useful)

And even if you had an explicit expression for y(x), you shouldn't expect to get zero when you plug in x = 0. You should expect to get y = 4 and y = 4/3 as solutions. Just because the value of the derivative of a function at a point is zero doesn't mean that the value of the function itself must be zero (in general...it could be in specific situation, eg for the function y = x^2, but there is no reason is has to be). The only restriction that the condition that y'(a) must be zero (at some point a) imposes is that the value of the function y(a) at that point must be an extremum. I hope that these corrections help. You seem to have some conceptual hang ups. I haven't checked your math over though...

Last edited: Mar 7, 2005
5. Mar 7, 2005

### erik05

Thanks cepheid for the explanation, it really helped clarify a few things and I hope my math is right.

Yes,I should really work on that....

Last edited: Mar 7, 2005