# Homework Help: Equation to the tangent

1. Jun 16, 2015

### says

1. The problem statement, all variables and given/known data
The cissoid of Diocles is given by the relation y2(2-x) = x3. Find the equation to the tangent line to the curve at the point (1,1).

2. Relevant equations

3. The attempt at a solution

Solution

d/dx [ y2(2-x) ] = d/dx [ x3 ]
2y dy/dx (2-x) + y2(-1) = 3x2
Therefore, dy/dx = 3x2+y2 / 2y(2-x)

m = dy/dx | (1,1) = 3+1/2(1) = 2 so y=2x+c and y(1) = 1, therefore 1=2+c ==> c=-1

Equation is y=2x-1

My attempt
y2(2-x) = x3
2y2-xy2-x2=0
d/dx [2y2-xy2-x3 = d/dx [0]
4y*dy/dx-y2-2xy*dy/dx-3x2 = 0
dy/dx [ 2xy - 4y ]-y2-3x2
dy/dx = -y2-3x2 / 2xy-4y

My solution gives m=1, therefore y=mx+c, 1=1*1+c, c=0 y=1...

The problem I'm having is that I don't understand why I can't expand the brackets in the original relation, y2(2-x) to be 2y2-xy2 or subtract x3 from both sides to make the equation 2y2-xy2-x2=0, which in my mind would make the problem easier to solve.

2. Jun 16, 2015

### SammyS

Staff Emeritus
You need to use adequate parentheses.
Your method also gives dy/dx = (-1 - 3(1) ) / ( 2(1)(1) - 4(1)) = (-4)/(-2) = 2

3. Jun 17, 2015

### says

@ (1,1)

dy/dx = (-y2-3x2)/( 2xy-4y)

dy/dx = (-12-3(-1)2) / (2(1)(1)-4(1))
= (1-3) / (2-4)
= -2 / -2
= 0

I see what I've done wrong. -1^2 = -1 ... but doesn't -1^2 = -1*-1 = 1 ???

4. Jun 17, 2015

### SammyS

Staff Emeritus
That's not (-y)2 , it's -(y2) .

5. Jun 17, 2015

### says

dy/dx = - (y2-3x2) / ( 2xy-4y)

like that?

6. Jun 17, 2015

### Staff: Mentor

No. -12 means - (12). If you want the square of -1, you need to write (-1)2

7. Jun 17, 2015

### says

ohhh ok. Is that the same for say, -x2... i.e. -x2 means -(x2)?

8. Jun 17, 2015

### Staff: Mentor

Yes

9. Jun 17, 2015

### Ray Vickson

You can do this; what makes you think otherwise?