Question

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:__________

Answer 1

Answer:

   -1    

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