Find al points where tangent lines to: (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x is either horizontal or vertical.
To find the points where the tangent lines to the curve are either horizontal or vertical, we first need to find the derivative of the curve. Let's begin by differentiating both sides of the equation:
Let's start by rewriting the equation as follows: (x^2 + y^2)^(3/2) - sqrt(x^2 + y^2) - x = 0
Now, we can differentiate both sides of the equation with respect to x:
d/dx [(x^2 + y^2)^(3/2) - sqrt(x^2 + y^2) - x] = 0
To differentiate the left side of the equation, we need to use the chain rule. Let's denote (x^2 + y^2) as u:
d/dx [(x^2 + y^2)^(3/2) - sqrt(x^2 + y^2) - x] = d/dx [u^(3/2) - sqrt(u) - x]
Using the chain rule, we can differentiate each term separately:
d/dx [u^(3/2)] = (3/2)u^(1/2) * d/dx [u] = (3/2)(x^2 + y^2)^(1/2) * (2x)
d/dx [sqrt(u)] = (1/2)(x^2 + y^2)^(-1/2) * d/dx [(x^2 + y^2)] = (1/2)(x^2 + y^2)^(-1/2) * (2x + 2yy')
d/dx [x] = 1
Now, let's substitute these derivatives back into our equation:
(3/2)(x^2 + y^2)^(1/2) * (2x) - (1/2)(x^2 + y^2)^(-1/2) * (2x + 2yy') - 1 = 0
Simplifying this equation, we have:
3x(x^2 + y^2)^(1/2) - (x^2 + y^2)^(-1/2)(x + y * dy/dx) - 1 = 0
Now, let's solve this equation for dy/dx, which represents the derivative of y with respect to x:
3x(x^2 + y^2)^(1/2) - (x^2 + y^2)^(-1/2)(x + y * dy/dx) - 1 = 0
Rearrange the equation to solve for dy/dx:
(x + y * dy/dx) / (x^2 + y^2)^(1/2) = 3x / (x^2 + y^2)^(1/2) - 1
Multiply both sides of the equation by (x^2 + y^2)^(1/2) to isolate dy/dx:
x + y * dy/dx = (3x - (x^2 + y^2)^(1/2)) * (x^2 + y^2)^(1/2)
Now, solve for dy/dx:
y * dy/dx = ((3x - (x^2 + y^2)^(1/2)) * (x^2 + y^2)^(1/2)) - x
Finally, divide both sides of the equation by y to get dy/dx:
dy/dx = ((3x - (x^2 + y^2)^(1/2)) * (x^2 + y^2)^(1/2) - x) / y
Now that we have the equation for dy/dx, we can set it equal to zero to find where the tangent line is horizontal, and find where it is undefined to find where the tangent line is vertical.