Equation for a tangent line?

Question

What would be the equation for a tangent line at x=0 if the curve's equation is x^3+3xy^2+y^3=1.

Answer 1

You'll need the tangent line's slope and intercept.

To find the slope, take the derivative dy/dx and evaluate it at x=0.
When x = 0, 0 + 0 + y^3 = 1, so y =1.

x^3 + 3xy^2 + y^3 = 1
(d/dx)(x^3 + 3xy^2 + y^3) = (d/dx)1
3x^2 + 3x2y(dy/dx) + 3y^2 + 3y^2(dy/dx) = 0
(3x^2 + 3y^2) + (dy/dx)(6xy + 3y^2) = 0
dy/dx = (6xy+3y^2)/(-3x^2 + 3y^2) = -(2xy + y^2)/(x^2 + y^2)
when x = 0, y = 1, and dy/dx = -1/1 = -1

Thus your slope is -1.

Your tangent line equation is thus y = -x + b.

To find the intercept, you know that the tangent contains the point x = 0, y = 1. Thus 1 = -0 + b, so b = 1

Thus the tangent line equation is y = -x + 1.

Answer 2

First find the point that the line is tangent at by plugging in x=0 to find that y^3 = 1 so the point is (0,1)

Then we need to find the slope at that point. Taking the derivative with respect to x gives:

3x^2 + 3y^2 + 6xyy' + 3y^2y' = 0

Now plug in x=0,y=1 to get:

3 + 3y' = 0 so y' = -1

So using the point-slope form (which in this case is the same as the slope-intercept form), we get the equation of the tangent line is:

y = -x + 1

Answer 3

Since you have multiple variables, you have to deal with partial derivates. First rewrite the equation as:

f(x,y) = x^3 + 3xy^2 + y^3 - 1

Find the partial derivates (for each variable's partial derivative, treat all other variables as constants and derive as you normally would):
fx(x,y) = 3x^2 + 3y^2
fy(x,y) = 6xy + 3y^2

By looking at the main equation , when x = 0, y = 1 obviously. So plug that into your partial derivative equations:
fx(0,1) = 3
fy(0,1) = 3

Those are the coefficients for the x and y in your tangent line equation. To find the constant in this linear equation, evalute it at (0,1) and set it equal to what your original equation evalutes to (1).
So now you have a tangent line equation as:
3x + 3y - 2 = 1
or
x + y = 1, which is a really nice equation!

Answer 4

The equation of tangent line can be given by derivative. All you have to do is differentiate the equation of curve with respect to x and you will get slope of tangent at given point.ie:-
d/dx(x^3+3xy^2+y^3=1)=3x^2+3[y^2+2xydy/dx]+3y^2dy/dx=0
3x^2+3y^2+6xydy/dx+3y^2dy/dx=0
x^2+y^2+2xydy/dx+y^2dy/dx=0
dy/dx=-(x^2+y^2)/(2xy+y^2)
Put x=0 in equation of curve to get value of y at that point where curve cuts y axis. we get:-y=1
Now put x=0,y=1 in above equation.
(dy/dx at x=0,y=1)=-1
This is slope of tangent at x=0, y=1:-
By slope point form:-
y-y1=m(x-x1)
y-1=-1(x-0)
y-1=-(1)x
y+x-1=0 is equation of tangent line