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Let xx and yy be functions of tt with y=4sinxy = 4\sin x. If dxdt=5\frac{dx}{dt} = 5, what is dydt\frac{dy}{dt} when x=π3x = \frac{\pi}{3}?\newlineWrite an exact, simplified answer.

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Q. Let xx and yy be functions of tt with y=4sinxy = 4\sin x. If dxdt=5\frac{dx}{dt} = 5, what is dydt\frac{dy}{dt} when x=π3x = \frac{\pi}{3}?\newlineWrite an exact, simplified answer.
  1. Identify Function and Differentiate: Identify the function and differentiate it with respect to xx.\newlineGiven y=4sin(x)y = 4\sin(x), differentiate to find dydx\frac{dy}{dx}.\newlinedydx=4cos(x)\frac{dy}{dx} = 4\cos(x)
  2. Use Chain Rule for dydt\frac{dy}{dt}: Use the chain rule to find dydt\frac{dy}{dt}. Given dxdt=5\frac{dx}{dt} = 5, use the chain rule: dydt=dydxdxdt\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}. dydt=4cos(x)5\frac{dy}{dt} = 4\cos(x) \cdot 5
  3. Substitute xx into dydt\frac{dy}{dt}: Substitute x=π3x = \frac{\pi}{3} into the equation.\newlineSubstitute x=π3x = \frac{\pi}{3} into dydt=4cos(x)×5\frac{dy}{dt} = 4\cos(x) \times 5.\newlinedydt=4cos(π3)×5\frac{dy}{dt} = 4\cos\left(\frac{\pi}{3}\right) \times 5\newlinedydt=4×(12)×5\frac{dy}{dt} = 4 \times \left(\frac{1}{2}\right) \times 5\newlinedydt=10\frac{dy}{dt} = 10

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