Find all theta in [0,2pi), where the curve: r(theta) = 6+4sin(theta) has a horizontal tangent. I got pi/2, 3pi/2, arcsin(-3/4) and arcsin(3/4)+pi, but they weren`t right. Any clues?
#Value of theta
summer garnet
cedar grove
Find all theta in [0,2pi), where the curve: r(theta) = 6+4sin(theta) has a horiz...
First, let's find out the expression of dy/dx in polar coordinates.
dx = (∂x/∂r)dr + (∂x/∂θ)dθ
dy = (∂y/∂r)dr + (∂y/∂θ)dθ
The partial derivatives are:
∂x/∂r = cos(θ)
∂x/∂θ = -r sin(θ)
∂y/∂r = sin(θ)
∂y/∂θ = r cos(θ)
So, substituting that and dividing dy by dx, we get:
dy/dx = (sin(θ)dr + r cos(θ)dθ)/(cos(θ)dr - r sin(θ)dθ)
Let's divide the numerator and denominator by θ.
dy/dx = ((dr/dθ)sin(θ) + r cos(θ))/((dr/dθ)cos(θ) - r sin(θ))
This is an expression we can use.
We need horizontal tangents, so dy/dx = 0. So:
(dr/dθ)sin(θ) + r cos(θ) = 0
(dr/dθ)cos(θ) - r sin(θ) ≠ 0
As r = 6 + 4sin(θ), dr/dθ = 4cos(θ). So:
4cos(θ)sin(θ) + (6 + 4sin(θ))cos(θ) = 0
4cos(θ)^2 - (6 + 4sin(θ))sin(θ) ≠ 0
Simplifying this a bit, we get:
(3 + 4sin(θ))cos(θ) = 0
2cos(2θ) - 3sin(θ) ≠ 0
Let's solve the first equation and then substitute the results into the second to verify that it is still an inequality.
(3 + 4sin(θ))cos(θ) = 0
Either:
cos(θ) = 0
θ = π/2 + πn
In our interval: θ = π/2, θ = 3π/2.
Or:
sin(θ) = -3/4
θ = -arcsin(3/4) + 2πn or θ = π + arcsin(3/4) + 2πn
In our interval: θ = π + arcsin(3/4), θ = 2π - arcsin(3/4)
You can verify that all these four points verify the inequality. So, the points are:
θ = π/2
θ = π + arcsin(3/4)
θ = 3π/2
θ = 2π - arcsin(3/4)
Seems you only missed θ = 3π/2.
summer garnet
Thank you so much. I had also the solution theta= 3pi/2. The problem was that I did arcsin(-3/4) instead of 2pi-arcsin(3/4). Regardless, your solution makes it much more simple now, thank you!
cedar grove
Thank you so much. I had also the solution theta= 3pi/2. The problem was that I ...
Ah, I see.
You're welcome!
How did you do it, by the way?
summer garnet
cedar grove
Ah, ok, same way, basically.
summer garnet
I learnt that you can do dy/dTheta= 0 instead of dy/dx as some say you can do (I haven`t implemented how to write math on discord yet).
cedar grove
I learnt that you can do dy/dTheta= 0 instead of dy/dx as some say you can do (I...
Hm... Let me check something.
(dr/dθ)sin(θ) + r cos(θ) = 0
(1/r)(dr/dθ) = -cot(θ)
(dr/dθ)cos(θ) - r sin(θ) ≠ 0
(1/r)(dr/dθ) ≠ tan(θ)
And I believe tan(θ) = -cot(θ) has no solutions.
So yeah, there isn't actually any need to verify that the denominator isn't equal to zero. The solutions of the first equation can never not be solutions of the inequality.