The equation x^2 y+2xy^3=8defines y as a function of x,y=f(x), near x=2, y=1.
a) Find dy/dx.
b) Now, find the slope of the curve
x^2y+2xy^3=8 when x=2, y=1.
differentiate implicitly,
x^2 + 2xy^3 = 8
2x + 2x(3y^2)dy/dx + 2(y^3) = 0
solve for dy/dx, then for b) sub in x=2, y=1
To find dy/dx, we can use implicit differentiation. Differentiating both sides of the equation with respect to x, we get:
d/dx[x^2y + 2xy^3] = d/dx[8]
Using the product rule, we can differentiate each term separately:
(d/dx[x^2y]) + (d/dx[2xy^3]) = 0
To differentiate each term, we treat y as a function of x and apply the chain rule:
2xy + x^2(dy/dx) + 2(1)(y^3) + 2x(3y^2)(dy/dx) = 0
Now our goal is to solve for dy/dx, since that represents the derivative dy/dx (the slope of the curve).
Rearranging the equation, we have:
x^2(dy/dx) + 2x(3y^2)(dy/dx) = -2xy - 2y^3
Factoring out dy/dx:
(dy/dx)(x^2 + 6xy^2) = -2xy - 2y^3
Now we can solve for dy/dx by dividing both sides of the equation by (x^2 + 6xy^2):
dy/dx = (-2xy - 2y^3) / (x^2 + 6xy^2)
To find the slope of the curve when x = 2 and y = 1, we substitute these values into the equation we just found:
slope = (-2(2)(1) - 2(1)^3) / (2^2 + 6(2)(1)^2)
simplify:
slope = (-4 - 2) / (4 + 6)
slope = -6 / 10
slope = -3/5
Therefore, the slope of the curve x^2y + 2xy^3 = 8 when x = 2 and y = 1 is -3/5.