How to: find the equation of the tangent line to the curve

curve: y^2=x^3(2-x) at the point (1,1)

here's my attempt:

2y*dy/dx = 3x^2(2-x)+x^3(-1)

2y*dy/dx = 6x^2-3x^3-x^3

2y*dy/dx = 6x^2 - 4x^3

dy/dx= (6x^2-4x^3)/2y

dy/dx = (6(1)^2-4(1)^3)/ 2(1)

=1

so for my answer I get

y-1=1(x-1)

y=x

is this correct? did I make any mistakes?

thanks!!

I was actually just doing some review of this..

lets use your example

y^2 = x^3*(2 - x)

we need the slope of the curve

2y*dy/dx = 3x^2*(2 -x) + x^3*(-1)

2y *dy/dx = 6x^2 - 3x^3 - x^3

2y * dy/dx = 6x^2 - 4x^3
dy/dx = (6x^2 - 4x^3)/2y

as you did you find the slope at the point

dy/dx = (6 - 4)/2 = 2/2 = 1

====================

lets create an equation for slope

y(a + dx) - y(a)
------------------- = slope
a + dx - a



y(a + dx) - y(a)
------------------- = slope
.......dx


solve for y(a + dx) this is the new point on the line

y(a + dx) - y(a) = slope * dx

also remember dx = x - a

I used a because its a point on the x graph that is constant.. so its less confusing.

y(a + dx) - y(a) = slope * (x - a)

y(a + dx ) = y(a) + slope*(x - a)

now we have the equation for the new point on the line
remember this is the TANGENT LINE only

Y = y(1) + slope*(x - 1)

the original equation y^2 = x^3*(2-x)

so y = sqrt(x^3 *(2-x))

put x = 1 into this

y = sqrt(1 * 1) = sqrt (1) = 1

Y = y(1) + slope*(x - 1)
Y = 1 + 1*(x - 1)

Y = 1 + x - 1 = x

so you were correct in the solution...