Calculus 1 question

Question

6xy+4x^3/2y^-1/2=56; (4,1) what is the slope of the curve

Answer 1

The slope is the first derivative (gradient) of the curve. differentiate the given curve using implicit and product rule.

6xdy/x +6y + 4x^3/2(-1/2)y^-3/2dy/dx + y^-1/24(3/2)x^1/2 = 0 [simplifying leads to]

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

==> slope = dy/dx = -3(x^1/2y^-1/2 + y)/(3x - 2x^3/2y^-3/2) [plug in the values of x = 4 and y =1 to get your slope]

Answer 2

6xy + 4x^3/2y^-1/2 = 56

6[y + x(dy/dx)] + 4[(3/2)x^1/2y^-1/2 + x^3/2(-1/2)y^-3/2(dy/dx)] = 0

Substituting x = 4 and y = 1, we have: 6[1 + 4(dy/dx)] + 4[(3/2)(2)(1) + 8(-1/2)(1)(dy/dx)] = 0

So, 6 + 24(dy/dx) + 12 - 16(dy/dx) = 0

8(dy/dx) = -18

dy/dx = -9/4