Question:

Consider the equation of the curve given by the implicit equation:

(x^2 + y^2)^3 = 8xy

a) Find dy/dx in terms of x and y using implicit differentiation.

b) Find the equation of the tangent line to the curve at the point (1, 1).

Solution:

a) To find the derivative of y with respect to x, we'll implicitly differentiate each term of the equation using the chain rule.

Given equation: (x^2 + y^2)^3 = 8xy

Step 1: Differentiate both sides of the equation with respect to x.

Using the chain rule, we have:

3(x^2 + y^2)^2 * (2x + 2yy') = 8y + 8xy'

Simplifying the equation, we get:

6(x^2 + y^2)^2x + 6(x^2 + y^2)^2yy' = 8y + 8xy'

Step 2: Rearrange the equation to solve for dy/dx.

6(x^2 + y^2)^2x - 8xy = 8y - 6(x^2 + y^2)^2yy'

Now, let's isolate dy/dx:

6(x^2 + y^2)^2yy' - 8xy = 8y - 6(x^2 + y^2)^2x

6(x^2 + y^2)^2yy' = 8y - 6(x^2 + y^2)^2x + 8xy

yy' = (8y - 6(x^2 + y^2)^2x + 8xy) / 6(x^2 + y^2)^2

dy/dx = [(8y - 6(x^2 + y^2)^2x + 8xy) / 6(x^2 + y^2)^2] / y

Therefore, the derivative dy/dx in terms of x and y is given by:

dy/dx = (8y - 6(x^2 + y^2)^2x + 8xy) / (6(x^2 + y^2)^2y)

b) To find the equation of the tangent line to the curve at the point (1, 1), we need to substitute these values into the derivative we just found.

Let's evaluate dy/dx at (1, 1):

dy/dx = (8(1) - 6(1^2 + 1^2)^2(1) + 8(1)(1)) / (6(1^2 + 1^2)^2(1))

Simplifying the expression:

dy/dx = (8 - 6(2)^2 + 8) / (6(2)^2) = (8 - 6(4) + 8) / (6(4)) = 4 / 24 = 1/6

The slope of the tangent line is 1/6.

Now, we can use the point-slope form of a linear equation to find the equation of the tangent line at (1, 1).

Using the point-slope form:

y - y1 = m(x - x1)

Substituting the point (1, 1) and the slope m = 1/6:

y - 1 = (1/6)(x - 1)

Expanding the equation:

6y - 6 = x - 1

Rearranging the equation:

x - 6y + 5 = 0

Therefore, the equation of the tangent line to the curve at the point (1, 1) is:

x - 6y + 5 = 0