find tangent and ppoints on curve where there is a horizontal or vertical tangent

Question

If I have the equation x³+3xy²+y³=1

 

What would be the equation for the tangent line when x=0, and what are all the points on the curve where the curve has a horizontal or vertical tangent? (I believe I would have to first find dy/dx in order to solve each of these problems, correct?) ?

Answer 1

Yes you would.  But since we have y in the function, we need to use implicit differentiation. 

 

Differentiate both sides of the equation using the chain rule.

 

3x² + 3y² + 6xyy' + 3y²y' = 0

 

 

Move all of the non-y' terms to the right side of equation.

 

6xyy' + 3y²y' = -3x² - 3y²

 

 

Factor out y'.

 

y' (6xy + 3y²) = -3x² - 3y²

 

 

y' = (-3x² - 3y²) / (6xy + 3y²)

 

 

We need to find the y-coordinate of the curve when x=0.  Plug in this x value into the curve given.

 

0³ + 3(0)(y²) + y³ = 1

 

y³ = 1

 

y = 1

 

 

So the point we are concentrating on is (0, 1).

 

Plug in these values into y'.

 

y' = -1

 

 

This is your slope of the tangent line.  Therefore, the equation of the tangent line is

 

y = -x + b

 

 

Substitute the value of the point into the equation to find b.

 

1 = -0 + b

 

1 = b

 

 

The equation of the tangent line is

 

y = -x + 1

 

 

To find the points on the curve where the curve has a horizontal tangent, we set the numerator part of y' equal to zero.  Solve for x.

 

To find the points on the curve where the curve has a vertical tangent, we set the denominator part y' equal to zero.  Solve for x.

 

 

Hope this helps!