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
[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{dy}{d \theta}}{\dfrac{dx}{d \theta}}[/tex]
[tex]\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}[/tex]
at θ = π/4 , r = sin π/2
r = 1
[tex]\dfrac{dy}{dx} = \dfrac{ r cos \theta + 2 cos (2 \theta)*sin \ \theta } {2 \ cos (2 \theta) *cos \theta- sin 2 \theta * sin \theta}[/tex]
where;
θ = π/4
[tex]\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})}[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{1}{\sqrt{2}}+0 }{0-1 * \dfrac{1}{\sqrt{2}}}[/tex]
[tex]\dfrac{dy}{dx} = \dfrac{\dfrac{1}{\sqrt{2}} }{- \dfrac{1}{\sqrt{2}}}[/tex]
[tex]\mathbf{\dfrac{dy}{dx} =-1}[/tex]
slope of the curve (dy/dx) at theta(θ) = pi/4 is -1
