Find the equation of the tangent line to the curve x 3 + xy 2 =5 at the point (1

Question

Find the equation of the tangent line to the curve x3 + xy2 =5 at the point (1,2). Question 2 options:
5x + 3y = 11
7x + 4y = 15
8x + 5y = 18
9x + 5y = 19
x + 3y = 7 Find the equation of the tangent line to the curve x3 + xy2 =5 at the point (1,2). Find the equation of the tangent line to the curve x3 + xy2 =5 at the point (1,2).
5x + 3y = 11
7x + 4y = 15
8x + 5y = 18
9x + 5y = 19
x + 3y = 7 5x + 3y = 11 7x + 4y = 15 8x + 5y = 18 9x + 5y = 19 x + 3y = 7
5x + 3y = 11
7x + 4y = 15
8x + 5y = 18
9x + 5y = 19
x + 3y = 7

Explanation / Answer

x^3+ xy^2 = 5

Take the derivative of this implicitly:

3x^2 + (x)(2y(dy/dx)) + (y^2)(1) = 0

3x^2 + 2xy(dy/dx) + y^2 = 0

Subtract 2xy(dy/dx) from each side:

3x ^2+ y^2 = -2xy(dy/dx)

Divide both sides by 2xy:

(3x^2 + y^2) / -2xy = dydx

Now plug in you point (1,2) to get your slope:

(3(1)^2 + (2)^2) / -2(1)(2)

= -7/4

So -7/4 will be the slope in this equation. Now plug in the y-value, x-value, and slope into y=mx+b form to solve for b:

2 = (-7/4)(1) + b

2 = (-7/4) + b

15/4 = b

so the end equation will look like this:

y = (-7/4)x + (15/4)

= B. 7x+4y = 15 <==Answer

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