Use implicit differentiation to find the slope of the tangent line to the trisectrix curve...

Use implicit differentiation to find the slope of the tangent line to the trisectrix curve y^(3)+yx^(2)+x^(2)−10y^(2)=0 at the point (0,10)

The word trisectrix may be intimidating, but in reality, all the question is asking you to do is two things:

1. Differentiate, that is, find the derivative of the equation given.

and

2. Find the value of the derivative at point (x,y) = (0,10).

Step 1. Finding the derivative:

Given equation:

y^3 + (x^2)y + x^2 - 10y^2 = 0

IMPLICITLY Differentiating—

(3y^2)(y') + 2xy + (x^2)(y') + 2x - (20y)(y') = 0 ..... [Remember to use product rule for the second term]

Grouping y' terms together, we get—

(3y^2)(y') + (x^2)(y') - (20y)(y') + 2x + 2xy = 0

y'(3y^2 + x^2 - 20y) + 2x + 2xy = 0 ..... Taking out y' as common multiple

y'(3y^2 + x^2 - 20y) = - 2x - 2xy ..... Subtracting both sides by 2x + 2xy

y' = (- 2x - 2xy) / (3y^2 + x^2 - 20y) ..... Dividing both sides by (3y^2 + x^2 - 20y)

We have successfully differentiated the given (trisectrix) curve's equation.

Step 2. Finding the value of this derivative at (x,y) = (0,10)

Plugging in x = 0; y = 10 in the derivative, we have—

y' = [-2(0) - 2(0)(10)] / [3(10)^2 + (0)^2 - 20 (10)]

Simplifying, we get—

y' = 0

Now, the derivative of a curve is the same as the slope of its tangent.

Therefore, y' is the same as m

Hence, m(slope) = 0

This is your WEBWORK answer.

Hope this helped!

y^(3)+yx^(2)+x^(2)−10y^(2)=0
3y^2dy/dx + x^2dy/dx +2xy +2x - 20ydy/dx = 0
At (0,10)
300dy/dx + 0 + 0 + 0 -200dy/dx = 0
100dy/dx = 0
dy/dx = 0
Slope = 0