Find the slope of the tangent line to the given polar curve at the point specified by the value of θ.
r=6/θ, θ=π
Why did the polar curve take up comedy? Because it wanted to find its tangent's "slope of humor!" Now, let's calculate it.
The given polar curve is r = 6/θ, and we need to find the slope of the tangent line at the point specified by θ = π. To do this, we need to find the derivative of r with respect to θ, and then substitute θ = π into that derivative.
Differentiating r = 6/θ with respect to θ, we have:
(dr/dθ) = -6/θ²
Now, we substitute θ = π:
(dr/dθ) evaluated at θ = π is:
(dr/dθ) = -6/(π)² = -6/π²
So, the slope of the tangent line to the given polar curve at the point specified by θ = π is -6/π². Keep up the great curiosity!
To find the slope of the tangent line to a polar curve at a specific point, we can use the formula:
slope = dy/dx = (dr/dθ * sin(θ) + r * cos(θ)) / (dr/dθ * cos(θ) - r * sin(θ))
Given the polar curve r = 6/θ and θ = π, let's find the slope.
1. First, let's find the derivative of r with respect to θ, dr/dθ:
Using the quotient rule, differentiate numerator and denominator separately.
dr/dθ = (d(6)/dθ * θ - 6 * d(θ)/dθ) / θ^2
= (0 * θ - 6 * 1) / θ^2
= -6 / θ^2
2. Next, let's find sin(θ) and cos(θ) at θ = π:
sin(π) = 0
cos(π) = -1
3. Substitute the values into the slope formula:
slope = (dr/dθ * sin(θ) + r * cos(θ)) / (dr/dθ * cos(θ) - r * sin(θ))
= (-6 / π^2 * 0 + (6/π) * -1) / (-6 / π^2 * -1 - (6/π) * 0)
= (0 + (-6/π)) / (6/π^2)
= -6π / 6
= -π
Therefore, the slope of the tangent line to the polar curve r = 6/θ at the point specified by θ = π is -π.
To find the slope of the tangent line to a polar curve at a specific point, we need to calculate the derivative of the curve with respect to θ and then evaluate it at the given value of θ.
In this case, we have the polar curve r = 6/θ and we want to find the slope of the tangent line at the point specified by θ = π.
First, let's calculate the derivative of the polar curve with respect to θ. To do this, we can use the chain rule. The chain rule states that if we have a function f(g(θ)), then the derivative of f(g(θ)) with respect to θ is given by f'(g(θ)) * g'(θ).
Let's consider f(g(θ)) = 6/g(θ), where g(θ) = θ. Taking the derivative of f(g(θ)), we get:
f'(g(θ)) = d/dθ (6/θ) = -6/θ²
Using the chain rule, we also need to find g'(θ), which is the derivative of θ with respect to θ. And that is simply 1.
So, the derivative of the polar curve r = 6/θ with respect to θ is given by:
d/dθ (r) = f'(g(θ)) * g'(θ) = (-6/θ²) * 1 = -6/θ²
Now, to find the slope of the tangent line at θ = π, we substitute π into our derivative:
slope = d/dθ (r) (θ = π) = (-6/π²)
Therefore, the slope of the tangent line to the given polar curve at the point specified by θ = π is -6/π².
y = r sinθ
x = r cosθ
dy = sinθ dr + r cosθ dθ
dx = cosθ dr - r sinθ dθ
so, the slope of the tangent line is
dy/dx = (sinθ dr + r cosθ dθ)/(cosθ dr - r sinθ dθ)
at θ=π, dy/dx = r dθ/dr = r(-6/r^2) = -π^2/6
so, now we have a point and a slope, so the tangent line is
y = -π^2/6 (x + 6/π)