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dmitriy555 [2]
3 years ago
6

Find the slope of the curve below at the given points. Sketch the curve along with its tangents at these points. R = sin 2 theta

: theta = plusminus pi/4, plusminus 3 pi/4 The slope of the curve at theta = pi/4 is:__________
Mathematics
1 answer:
mr Goodwill [35]3 years ago
7 0

Answer:

<u>    </u><u>-1     </u>

Step-by-step explanation:

Given that:

r = sin 2θ  , θ = ± π/4 , ± 3π/4

Recall that:

x = r cosθ

y = r sinθ

The  differential of y with respect to x

\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}}

\dfrac{dy}{dx} =\dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}} = \dfrac{ r cos \theta + sin \theta* \dfrac{dr}{d \theta} } {\dfrac{dr}{d \theta} *cos \theta- sin \theta \ r}

at  θ = π/4  , r = sin π/2

r = 1

\dfrac{dy}{dx} = \dfrac{ r cos \theta + 2 cos (2 \theta)*sin \ \theta } {2 \ cos (2 \theta) *cos \theta- sin 2 \theta * sin \theta}

where;

θ = π/4

\dfrac{dy}{dx} = \dfrac{ 1 \times  cos (\dfrac{\pi}{4}) + 2 cos  (\dfrac{\pi}{2})*sin  (\dfrac{\pi}{4}) } {2 \ cos  (\dfrac{\pi}{2})*cos  (\dfrac{\pi}{4})- sin  (\dfrac{\pi}{2}) * sin  (\dfrac{\pi}{4})}

\dfrac{dy}{dx} = \dfrac{\dfrac{1}{\sqrt{2}}+0 }{0-1 * \dfrac{1}{\sqrt{2}}}

\dfrac{dy}{dx} = \dfrac{\dfrac{1}{\sqrt{2}} }{- \dfrac{1}{\sqrt{2}}}

\mathbf{\dfrac{dy}{dx} =-1}

slope of the curve (dy/dx) at theta(θ)  = pi/4 is -1

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