Find the slope dy/dx of the polar curve r = 1 + 2 sin θ at θ = pi over 2
a) Undefined
b) 1
c) -1
d) 0
well, you know it is a limaçon with a vertical axis, so it has a horizontal tangent at θ=π/2
To confirm that, recall that
y = rsinθ
x = rcosθ
dy/dθ = r' sinθ + rcosθ
dx/dθ = r'cosθ - rsinθ
r' = 2cosθ
so, at θ = π/2, r=3, and we have
dy/dθ = 0*1 + 3*0 = 0
dx/dθ = 0*0 - 3*1 = -3
dy/dx = (dy/dθ) / (dx/dθ) = 0/-3 = 0
To find the slope dy/dx of a polar curve, we can use the formula:
dy/dx = (dr/dθ sin θ + r cos θ) / (dr/dθ cos θ - r sin θ)
In this case, the polar curve is r = 1 + 2 sin θ and we need to find the slope at θ = π/2.
First, we need to find dr/dθ. To do this, we differentiate r with respect to θ:
dr/dθ = 2 cos θ
Next, we substitute the values into the slope formula:
dy/dx = (2 cos θ sin θ + (1 + 2 sin θ) cos θ) / (2 cos θ cos θ - (1 + 2 sin θ) sin θ)
At θ = π/2, cos θ = 0 and sin θ = 1. Substituting these values into the equation:
dy/dx = (2(0)(1) + (1 + 2(1))(0)) / (2(0)(0) - (1 + 2(1))(1))
= 0 / (0 - (1 + 2))
= 0 / -3
= 0
Therefore, the slope dy/dx of the polar curve r = 1 + 2 sin θ at θ = π/2 is 0.
The answer is (d) 0.
To find the slope of the polar curve \(r = 1 + 2\sin(\theta)\), we need to find the derivative \(\frac{dy}{dx}\) using polar coordinates. The derivative in polar coordinates is given by the formula:
\[\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}\]
To find the slope at \(\theta = \frac{\pi}{2}\), we need to substitute \(\theta = \frac{\pi}{2}\) into the equation and evaluate.
First, let's find \(\frac{dr}{d\theta}\):
\[r = 1 + 2\sin(\theta)\]
\[\frac{dr}{d\theta} = 2\cos(\theta)\]
Now, we substitute \(\theta = \frac{\pi}{2}\) into the equation:
\[\frac{dy}{dx} = \frac{\frac{dr}{d\theta}\sin(\theta) + r\cos(\theta)}{\frac{dr}{d\theta}\cos(\theta) - r\sin(\theta)}\]
\[\frac{dy}{dx} = \frac{2\cos(\frac{\pi}{2})\sin(\frac{\pi}{2}) + (1 + 2\sin(\frac{\pi}{2}))\cos(\frac{\pi}{2})}{2\cos(\frac{\pi}{2})\cos(\frac{\pi}{2}) - (1 + 2\sin(\frac{\pi}{2}))\sin(\frac{\pi}{2})}\]
\[\frac{dy}{dx} = \frac{2(0)(1) + (1 + 2)(0)}{2(0)(0) - (1 + 2)(1)}\]
\[\frac{dy}{dx} = \frac{0 + 0}{0 - 3}\]
Since we have a zero in the numerator and a negative number in the denominator, the expression is undefined. Therefore, the slope \(\frac{dy}{dx}\) of the polar curve \(r = 1 + 2\sin(\theta)\) at \(\theta = \frac{\pi}{2}\) is undefined (choice a).