What Is the Equation of the Tangent to the Cissoid of Diocles at (1,1)?

[says]

Homework Statement

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).

Homework Equations

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.

 

[SammyS]

Homework Statement

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).

Homework Equations

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

You need to use adequate parentheses.

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.

Your method also gives dy/dx = (-1 - 3(1) ) / ( 2(1)(1) - 4(1)) = (-4)/(-2) = 2

 

[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 ?

 

[SammyS]

@ (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 ?

That's not (-y)² , it's -(y²) .

 

[says]

dy/dx = - (y²-3x²) / ( 2xy-4y)

like that?

 

[Mark44]

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

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

 

[says]

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

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

 

[Mark44]

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

Yes

 

[Ray Vickson]

Homework Statement

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).

Homework Equations

The Attempt at a Solution

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.

You can do this; what makes you think otherwise?