Find Equations of Tangents to C1 and C2 | Area Enclosed | Derivatives

[songoku]

Homework Statement

let C₁ : y = x - 1/2 x² and C₂ : x = y - 1/2 y² be curves on the xy plane.

1. find the equation of the tangent to the curve C₁ at x = k

2. suppose the line obtained in 1) is also tangent to the curve C₂. find all values of k and the equations of the tangents.

3. evaluate the area of the figure enclosed by all tangents obtained in 2) and the curve C₂

Homework Equations

derivatives, equation of tangent

The Attempt at a Solution

1. I've done it. I got : y = (1-k) x + 1/2 k²

2.
differentiate C₂ with respect to x :
1 = dy/dx - y dy/dx

dy/dx = 1/(1-y) = m

so, 1/(1-y) = 1-k

then...I...gave up...

 

[rock.freak667]

If x=k is a point on C₁, then the point (k,k-1/2k²) ((x,y)) is a point on the line right?


So the gradient of the tangent on C₂ is dy/dx= 1/(1-y). Since (k,k-1/2k²) lies on the line, what is the gradient of this tangent?


then put that equal to 1-k.

 

[songoku]

If x=k is a point on C₁, then the point (k,k-1/2k²) ((x,y)) is a point on the line right?


So the gradient of the tangent on C₂ is dy/dx= 1/(1-y). Since (k,k-1/2k²) lies on the line, what is the gradient of this tangent?


then put that equal to 1-k.

Do you mean substituting y = k - 1/2 k² to dy/dx= 1/(1-y) ?

 

[annoymage]

sorry (editted)

alternative way,

you also can find another tangent equation with gradient 1/(1-y) with point (k,k-1/2k2), and compare with the other tangent eqution

 

[songoku]

sorry (editted)

alternative way,

you also can find another tangent equation with gradient 1/(1-y) with point (k,k-1/2k2), and compare with the other tangent eqution

I don't get it. x = k is the common point between the tangent and C₁ and I think we can't use it to find the equation of tangent of C₂ since x = k may not be the common point.

Or maybe I missed the hint?

 

[annoymage]

it says that, the tangent of C1 at x=k , is also tangent C2 at some point x, implies that the tangent of C1 and tangent of C2 is the same line.

since it is the same line, tangent C2 also pass through point (k,k-1/2k2)

 

[annoymage]

i'm sorry but i think I'm wrong