Find al points where tangent lines to: (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x is either horizontal or vertical using implicit differentiation.
3√(x^2+y^2) * (x + yy') = (x+yy')/√(x^2+y^2) + 1
x+yy'(3√(x^2+y^2) - 1/√(x^2+y^2) = 1
y' = (1/(3√(x^2+y^2) - 1/√(x^2+y^2) - x)/y
y' = [x+√(x^2+y^2) - 3x(x^2+y^2)]/(y(3(x^2+y^2)-1))
So, we want either the numerator or the denominator to be zero.
It might be a bit easier to use polar coordinates.
y' = (x+r-3xr^2)/(3yr^2-y)
= (rcosθ+r-3r^3 cosθ)/(rsinθ(3r^2-1))
r^2 = 1+cosθ
The tangent is vertical where
rsinθ(3r^2-1) = 0
or, at θ = 0 or 2pi
the tangent is horizontal when
r(cosθ+1-3r^2cosθ) = 0
r(cosθ+1-3cosθ(1+cosθ)) = 0
cosθ+1-3cosθ-3cos^2θ = 0
3cos^2θ + 2cosθ - 1 = 0
(3cosθ-1)(cosθ+1) = 0
So, horizontal when
cosθ = -1
cosθ = 1/3
Better double-check the math
Thank you!
To find the points where the tangent lines to the curve (x^2+y^2)^(3/2) = sqrt(x^2+y^2) + x are either horizontal or vertical, we can use implicit differentiation.
Implicit differentiation involves differentiating both sides of the equation with respect to x, treating y as a function of x and using the chain rule.
Let's begin by differentiating both sides of the equation with respect to x.
For the left-hand side, we apply the chain rule:
d/dx[(x^2+y^2)^(3/2)] = d/dx[(x^2+y^2)^(3/2)]*(x^2+y^2)^(1/2) *d/dx(x^2+y^2)
For the right-hand side, we differentiate each term individually:
d/dx[sqrt(x^2+y^2) + x] = d/dx[sqrt(x^2+y^2)] + d/dx[x]
Let's simplify each term separately:
Starting with the left-hand side:
d/dx[(x^2+y^2)^(3/2)] = (3/2)*(x^2+y^2)^(1/2) * (2x + 2y * dy/dx)
Next, let's differentiate the right-hand side:
d/dx[sqrt(x^2+y^2)] = (1/2)*(x^2+y^2)^(-1/2) * (2x + 2y * dy/dx)
d/dx[x] = 1
Combining these results, we have:
(3/2)*(x^2+y^2)^(1/2) * (2x + 2y * dy/dx) = (1/2)*(x^2+y^2)^(-1/2) * (2x + 2y * dy/dx) + 1
Now, let's find the points where the slope of the tangent line is either horizontal or vertical.
For a horizontal tangent line, the slope is equal to 0.
So we set dy/dx = 0 and solve for the corresponding points.
(3/2)*(x^2+y^2)^(1/2) * (2x) = (1/2)*(x^2+y^2)^(-1/2) * 2x + 1
Simplifying further:
3*(x^2+y^2) = (x^2+y^2)^(-1/2) + (x^2+y^2)^(1/2) + (x^2+y^2)
(3*(x^2+y^2))^2 = 1 + 2 * (x^2+y^2)^(-1/2) + (x^2+y^2)
9(x^2+y^2)^2 = 1 + 2/(x^2+y^2) + (x^2+y^2)
This is a fourth-degree equation in terms of x and y. By solving this equation, you can find the points where the slope of the tangent line is horizontal.
For a vertical tangent line, the slope is undefined (denominator equals 0).
So, we set the denominator of the derivative expression equal to 0 and solve for the corresponding points.
(x^2+y^2)^(1/2) = 0
This equation simplifies to:
x^2 + y^2 = 0
Since x^2 and y^2 are both non-negative, the only solution to this equation is the point (0,0).
To summarize, to find the points where the tangent lines to the given curve are either horizontal or vertical using implicit differentiation:
1. Set dy/dx = 0 to find the points where the slope is horizontal.
2. Set the denominator of the derivative expression equal to 0 to find the points where the slope is vertical.
3. Solve the resulting equations to obtain the specific points.