(dr)/(dθ)=8(cos(θ/3))^3

Lösung

\frac{d r}{d θ} = 8 (cos (\frac{θ}{3}))^{3}

r = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

Schritte zur Lösung

\frac{d r}{d θ} = 8 (cos (\frac{θ}{3}))^{3}

Ersetze

\frac{d r}{d θ}

mit

r^{'} (θ)

r^{^{'}} (θ) = 8 (cos (\frac{θ}{3}))^{3}

Wenn

f^{^{'}} (x) = g (x)

dann

f (x) = \int g (x) d x

r = \int 8 (cos (\frac{θ}{3}))^{3} d θ

\int 8 (cos (\frac{θ}{3}))^{3} d θ = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

r = 24 (sin (\frac{θ}{3}) - \frac{1}{3} sin^{3} (\frac{θ}{3})) + c_{1}

x \to 0 + lim ((e^{x} + x)^{\frac{2}{x}})

\frac{d}{d x} ((\frac{x + 1}{x - 1})^{\frac{1}{2}})

\int \frac{1}{3 x ^{2} + 4 x + 5} d x

\frac{d}{d x} (- a x)

\frac{\partial}{\partial x} (cos (8 x) sin (3 y))