Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = 1 + 2 cos θ, θ = π/3

Find the slope of the tangent line to the given polar curve at the point specified by the value of θ. r = 1 + 2 cos θ, θ = π/3

Expert Solution

Solution

step

y = (1 + 2 cos θ) \cdot sin θ

Differentiate with respect to $\theta$ by using the product rule

\dfrac{dy}{d\theta}=\left(0-2\sin\theta\right)\cdot\sin\theta+\left(1+2\cos\theta\right)\cdot\left(\cos\theta\right)

Substitute $\theta=\pi/3$ (which means $\sin\theta=\sqrt{3}/2$ and $\cos\theta=1/2$ )

\dfrac{dy}{d\theta}\bigg|_{\theta=\pi/3}=\left(0-2\cdot\dfrac{\sqrt{3}}{2}\right)\cdot\dfrac{\sqrt{3}}{2}+\left(1+2\cdot\dfrac{1}{2}\right)\cdot\left(\dfrac{1}{2}\right)

\dfrac{dy}{d\theta}\bigg|_{\theta=\pi/3}=\left(-\sqrt{3}\right)\cdot\dfrac{\sqrt{3}}{2}+\left(1+1\right)\cdot\left(\dfrac{1}{2}\right)

\dfrac{dy}{d\theta}\bigg|_{\theta=\pi/3}=-\dfrac{3}{2}+1=-\dfrac{3}{2}+\dfrac{2}{2}=-\dfrac{1}{2}

Using the chain rule, we can write

\dfrac{dy}{dx}\bigg|_{\theta=\pi/3}=\dfrac{dy/d\theta}{dx/d\theta}\bigg|_{\theta=\pi/3}=\dfrac{-1/2}{-3\sqrt{3}/2}=\dfrac{1}{3\sqrt{3}}

Slope of the tangent is $\dfrac{1}{3\sqrt{3}}$

SUN BUNRA answered 2 years ago

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