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Question

can u help me on this too??

Find the slope of the curve (x^2+y^2)^2=4y^2 at the point (0,2)...(using implicit differentiation).

Answer 1

(x^2 + y^2)^2 = 4y^2

The first step is to implicitly differentiate. Remember the chain rule, and remember the derivative of y is dy/dx. On the left hand side, we're going to apply the power rule, then chain rule, and then chain rule on the y^2. On the right hand side, it's just the power rule and the chain rule.

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

Let's distribute 2(x^2 + y^2) over the second set of brackets.

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

And some more simplification...
4x(x^2 + y^2) + 4y(x^2 + y^2)(dy/dx) = 8y (dy/dx)

Now, we move everything with a (dy/dx) on the left hand side, and everything else on the right hand side.

4y(x^2 + y^2)(dy/dx) - 8y(dy/dx) = -4x(x^2 + y^2)

Factor (dy/dx) out of the terms,

(dy/dx) [4y(x^2 + y^2) - 8y] = -4x(x^2 + y^2)

And then divide both sides to isolte (dy/dx)...

dy/dx = [-4x(x^2 + y^2)] / [4y(x^2 + y^2) - 8y]

To find the slope of the curve at (0,2), all we do is plug in x=0 and y=2. Notice that since we have -4x on the numerator as a product, assigning x = 0 makes everything immediately 0.

Therefore, the slope of the curve at the point (0,2) is 0.

Answer 2

That's not too hard. Just use implicit differentiation as you said, and substitute 0 for x and 2 for y, and solve for y prime.

Answer 3

(x^2 + y^2)^2 = 4y^2
2(x^2 + y^2)[2x + 2y(dy/dx)] = 8y(dy/dx)
4x^3 + 4(x^2)y(dy/dx) + 4x(y^2) + 4(y^3)(dy/dx) = 8y(dy/dx)
x^3 + (x^2)y(dy/dx) + x(y^2) + (y^3)(dy/dx) = 2y(dy/dx)
(dy/dx)[(x^2)y + y^3 - 2y] = -x^3 - x(y^2)
(dy/dx) = -[x^3 + x(y^2)]/[(x^2)y + y^3 - 2y]

Plugging in (0,2) for (x,y) we get

(dy/dx) = [0 + 0]/[0 + 8 - 4] = 0