Find the tangent to the curve x3 y3 6xy at the point 3 3So

Find the tangent to the curve x^3 + y^3 = 6xy at the point (3, 3).

Solution:

y = x + 6 is the tangent to the curve.

Explanation:

to find the equation of the tangent to a curve by the point-slope method.

x ³ + y ³ = 6xy

Differentiating wrt x we have

3x ² + 3y ² dy /dx= 6y + 6x dy/ dx

{(the formulas used above is dy/dx(x^{n )}=nx^n-1

d/dx(x)=1 and RHS= d/dx(uv=uv¹ +vu¹)}

3y ² dy /dx- 6x dy/ dx= 6y-3x ²

dy /dx(3y ² -6x )= 6y-3x ²

dy /dx*3(y ² -2x )=3(2y-x ² )

cancel 3 on both sides we get

dy/dx=(2y x ²) /(y ² 2x)------(A)

dy/dx is called slope of line represented by m

put x=3 and y=3 in(A) to get the slope,m

m=dy/dx=2*3-3²/3²-2(3)

m=6-9/9-6

=-3/3

m=-1

Therfore,we find that the slope at (3, 3) is 1

Using point slope form of line we have

y-y₁ =m(x-x₁)

we have (x1,y1)=(3,3) and slope=m=-1

Therefore ,the corresponding tangent has equation y 3 = (1)(x 3)

or y = x + 6.