for what value of r=4sinè have vertical tangent?
In polar coordinates? Theta=90 deg is vertical.
I do not know if you know how to differentiate polar coordinate equations so I changed your equation from polar to rectangular form, I don't know if that is the way you do it.
anyway I got x^2 + y^2 = 4y which is the equation of a circle.
I differentiated that
2x + 2y(dy/dt) = 4dy/dt
.
.
dy/dt = -x/(y-2)
for the tangent to be a vertical line, its slope must be undefined.
This happens when the denominator is 0
so y-2=0
y=2
then substitute that in the equation for x=2
so in Cartesian form the point is (2,2), making an angle of 45º or pi/4 radians.
in polar form (r,theta) it would be (√8,pi/4)
To find the value of r for which the equation r = 4sin(θ) has a vertical tangent, we can use polar coordinates. In polar coordinates, a vertical tangent occurs when θ = 90 degrees (or π/2 radians).
First, let's rewrite the equation in rectangular form. The equation r = 4sin(θ) represents a circle centered at the origin with radius 4.
To convert the equation to rectangular form, we can use the relationship x = rcos(θ) and y = rsin(θ). Substituting these values, we get x^2 + y^2 = (4sin(θ))^2.
Simplifying further, we have x^2 + y^2 = 16sin^2(θ).
Next, let's differentiate the equation with respect to θ to find the slope of the tangent line. Using implicit differentiation, we get:
2x(dx/dθ) + 2y(dy/dθ) = 32sin(θ)cos(θ).
Now, we need to find the value of θ for which the slope of the tangent line is undefined (a vertical tangent). This occurs when the denominator of dy/dθ becomes zero.
Setting the denominator equal to zero, we have y - 2 = 0, which gives y = 2.
Substituting y = 2 back into the original equation x^2 + y^2 = 16sin^2(θ), we get x^2 + 4 = 16sin^2(θ).
Simplifying further, x^2 = 16sin^2(θ) - 4.
Now, substitute y = 2 and solve for x. We have x^2 + 2^2 = 16sin^2(θ) - 4, which leads to x^2 + 4 = 16sin^2(θ).
Since sin^2(θ) can be represented as (1 - cos^2(θ)), we have x^2 + 4 = 16(1 - cos^2(θ)).
Simplifying further, we get x^2 + 4 = 16 - 16cos^2(θ).
Rearranging the equation, we have x^2 = 12 - 16cos^2(θ).
Substituting θ = 90 degrees (or π/2 radians) to find the vertical tangent, we get x^2 = 12 - 16 * 0.
Simplifying further, x^2 = 12.
Taking the square root of both sides, we have x = ±√12, which simplifies to x = ±2√3.
Therefore, in Cartesian form, the point where the equation r = 4sin(θ) has a vertical tangent is (2√3, 2).
Converting this point back to polar form, we have (r, θ) = (√(2^2 + 2^2), arctan(2/2√3)).
Simplifying, we get (r, θ) = (√8, π/4).
So, the value of r for which the equation r = 4sin(θ) has a vertical tangent is √8 when θ = π/4 radians.