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A curve is defined by the parametric equations x(t)=-3cos(9t) and y(t)=-9e^(-10 t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=−3cos⁡(9t) x(t)=-3 \cos (9 t) and y(t)=−9e−10t y(t)=-9 e^{-10 t} . Find dydx \frac{d y}{d x} .\newlineAnswer:

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Q. A curve is defined by the parametric equations x(t)=−3cos⁡(9t) x(t)=-3 \cos (9 t) and y(t)=−9e−10t y(t)=-9 e^{-10 t} . Find dydx \frac{d y}{d x} .\newlineAnswer:
  1. Find dxdt\frac{dx}{dt}: To find dydx\frac{dy}{dx} for parametric equations, we need to find dydt\frac{dy}{dt} and dxdt\frac{dx}{dt} separately and then divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}.
  2. Find dydt\frac{dy}{dt}: First, we find dxdt\frac{dx}{dt}. The derivative of x(t)=−3cos⁡(9t)x(t) = -3\cos(9t) with respect to tt is dxdt=ddt[−3cos⁡(9t)]=27sin⁡(9t)\frac{dx}{dt} = \frac{d}{dt} [-3\cos(9t)] = 27\sin(9t).
  3. Divide to find dydx\frac{dy}{dx}: Next, we find dydt\frac{dy}{dt}. The derivative of y(t)=−9e−10ty(t) = -9e^{-10t} with respect to tt is dydt=ddt[−9e−10t]=90e−10t\frac{dy}{dt} = \frac{d}{dt} [-9e^{-10t}] = 90e^{-10t}.
  4. Simplify dydx\frac{dy}{dx}: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}. So, dydx=dydtdxdt=90e−10t27sin⁡(9t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{90e^{-10t}}{27\sin(9t)}.
  5. Final answer: We simplify the expression for dydx\frac{dy}{dx} to get the final answer. dydx=9027e−10t/sin⁡(9t)=103e−10t/sin⁡(9t)\frac{dy}{dx} = \frac{90}{27}e^{-10t} / \sin(9t) = \frac{10}{3}e^{-10t} / \sin(9t).

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