Find the slope of the tangent to the curve x^3+7xy+y^2=19 at (1,2). The slope =?

Answer 1

Given; y = x^3 + 7xy + y^2 = 19

d/dx [ x^3 + 7xy + y^2 ] = d/dx [19]

3x^2 + 7{ y + xy' } + 2yy' = 0

3x^2 + 7y + 7xy' + 2yy' = 0

Solve for y' in terms of x and y

7xy' + 2yy' = - 3x^2 - 7y

y'(7x + 2y) = - 3x^2 - 7y

y ' = [ -3x^2 - 7y ] / [ 7x + 2y ]

When: (x,y) = (1,2)

y ' = [ -3x^2 - 7y ] / [ 7x + 2y]

y ' = [ -3 - 14 ] / [ 7 + 4 ]

y ' = [-17] / [ 11 ]

y' = -17/11

CHECK: (x,y) = (1,2) and y' = -17/11

3x^2 + 7y + 7xy' + 2yy' = 0

3(1)^2 + 7(2) - 7(1)(17/11) - 2(2)(17/11) = 0

3 + 14 - 119/11 - 68/11 = 0

17 - 187/11 = 0

17 - 17 = 0

0 = 0 TRUE

Answer 2

Use implicit differentiation to find dy/dx (or y'), which is the slope of the tangent line

Take the derivative of both sides, with respect to x

3x^2 + 7x y' + 7y + 2y y' = 0

Solve for y'

7x y' + 2y y' = -3x^2 - 7y

y'(7x + 2y) = -3x^2 - 7y

y' = (-3x^2 - 7y)/(7x + 2y)

Now, substitute x = 1 and y = 2

y' = (-3(1)^2 - 7(2))/(7(1) + 2(2)) = -17/11.

So, the slope of the tangent line at (1,2) is -17/11