Exercício 4: Calcule as expressões das derivadas e os seus respectivos valores nos pontos dados:
\left\{\begin{array}{l} x=\operatorname{sen} t \\\ y=\operatorname{sen} 2 t \end{array}, t \in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right], \frac{d y}{d x}, \text { no ponto } t=\frac{\pi}{6}\right.
\left\{\begin{array}{l} x=6 t\left(1+t^{2}\right)^{-1} \\\ y=6 t^{2}\left(1+t^{2}\right)^{-1} \end{array}, 0 \leq t \leq 1, \frac{d y}{d x}, \text { no ponto de abscissa } \frac{12}{5}\right.
\left\{\begin{array}{l} x=t+\operatorname{sen}\left(\frac{\pi}{2} t\right) \\\ y=t+\ln t \end{array}, t>0, \frac{d y}{d x}, \text { no ponto } t=8\right.

Passo 1

Temos três funções paramétricas (x(t),y(t)) e queremos calcular
\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}
em cada um dos pontos indicados. Vamos derivar x(t) e y(t) em função de t, montar a razão \frac{dy/dt}{dx/dt} e avaliar no t pedido.

Passo 2

Para
x=\operatorname{sen} t,\quad y=\operatorname{sen}(2t),\quad t\in\Bigl[-\tfrac\pi2,\tfrac\pi2\Bigr],
temos
\frac{dx}{dt}=\cos t, \qquad \frac{dy}{dt}=2\cos(2t).
Logo
\frac{dy}{dx} =\frac{2\cos(2t)}{\cos t}.
No ponto t=\frac\pi6 calculamos
\cos\Bigl(2\cdot\frac\pi6\Bigr)=\cos\frac\pi3=\tfrac12, \qquad \cos\frac\pi6=\tfrac{\sqrt3}{2},
então
\left.\frac{dy}{dx}\right|_{t=\pi/6} =\frac{2\cdot\frac12}{\frac{\sqrt3}{2}} =\frac{1}{\tfrac{\sqrt3}{2}} =\frac{2}{\sqrt3} =\frac{2\sqrt3}{3}.

Passo 3

Para
x=\frac{6t}{1+t^2},\quad y=\frac{6t^2}{1+t^2},\quad 0\le t\le1,
usamos a regra do quociente:
\frac{dx}{dt} =\frac{6(1+t^2)-6t\cdot2t}{(1+t^2)^2} =\frac{6-6t^2}{(1+t^2)^2} =\frac{6(1-t^2)}{(1+t^2)^2},
\frac{dy}{dt} =\frac{12t(1+t^2)-6t^2\cdot2t}{(1+t^2)^2} =\frac{12t}{(1+t^2)^2}.
Logo
\frac{dy}{dx} =\frac{\frac{dy}{dt}}{\frac{dx}{dt}} =\frac{12t}{6(1-t^2)} =\frac{2t}{1-t^2}.
Precisamos do ponto em que x=\tfrac{12}{5}. Resolvendo
\frac{6t}{1+t^2}=\frac{12}{5} \quad\Longrightarrow\quad 30t=12+12t^2 \quad\Longrightarrow\quad 2t^2-5t+2=0
temos t=\tfrac12 (o outro t=2 está fora de [0,1]). Assim
\left.\frac{dy}{dx}\right|_{x=12/5} =\left.\frac{2t}{1-t^2}\right|_{t=1/2} =\frac{2\cdot\frac12}{1-\frac14} =\frac{1}{\tfrac34} =\frac{4}{3}.

Passo 4

Para
x=t+\operatorname{sen}\Bigl(\tfrac{\pi}{2}t\Bigr), \quad y=t+\ln t, \quad t>0,
derivamos
\frac{dx}{dt} =1+\cos\Bigl(\tfrac{\pi}{2}t\Bigr)\cdot\frac{\pi}{2}, \qquad \frac{dy}{dt} =1+\frac1t.
Portanto
\frac{dy}{dx} =\frac{1+\frac1t}{1+\frac{\pi}{2}\cos\bigl(\tfrac{\pi}{2}t\bigr)}.
No ponto t=8 temos \cos(\tfrac{\pi}{2}\cdot8)=\cos(4\pi)=1, logo
\left.\frac{dy}{dx}\right|_{t=8} =\frac{1+\frac{1}{8}}{1+\frac{\pi}{2}\cdot1} =\frac{\tfrac{9}{8}}{1+\tfrac{\pi}{2}} =\frac{9}{8}\cdot\frac{2}{2+\pi} =\frac{9}{4(2+\pi)}.

Resposta

Para x=\operatorname{sen} t,\;y=\sen2t em t=\tfrac\pi6:
\frac{dy}{dx}=\frac{2\sqrt3}{3}.
Para x=\tfrac{6t}{1+t^2},\;y=\tfrac{6t^2}{1+t^2} com abscissa x=\tfrac{12}{5}:
\frac{dy}{dx}=\frac{4}{3}.
Para x=t+\operatorname{sen}(\tfrac{\pi}{2}t),\;y=t+\ln t em t=8:
\frac{dy}{dx}=\frac{9}{4(2+\pi)}.

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