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A curve is defined by the parametric equations x(t)=-4cos(-5t) and y(t)=3sin(-8t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=−4cos⁡(−5t) x(t)=-4 \cos (-5 t) and y(t)=3sin⁡(−8t) y(t)=3 \sin (-8 t) . Find dydx \frac{d y}{d x} .\newlineAnswer:

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Q. A curve is defined by the parametric equations x(t)=−4cos⁡(−5t) x(t)=-4 \cos (-5 t) and y(t)=3sin⁡(−8t) y(t)=3 \sin (-8 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)=−4cos⁡(−5t)x(t) = -4\cos(-5t) with respect to tt is dxdt=−4⋅ddt[cos⁡(−5t)]\frac{dx}{dt} = -4 \cdot \frac{d}{dt}[\cos(-5t)]. Using the chain rule, we get dxdt=−4⋅(−5)⋅(−sin⁡(−5t))=20sin⁡(−5t)\frac{dx}{dt} = -4 \cdot (-5) \cdot (-\sin(-5t)) = 20\sin(-5t).
  3. Divide to find dydx\frac{dy}{dx}: Next, we find dydt\frac{dy}{dt}. The derivative of y(t)=3sin⁡(−8t)y(t) = 3\sin(-8t) with respect to tt is dydt=3⋅ddt[sin⁡(−8t)]\frac{dy}{dt} = 3 \cdot \frac{d}{dt}[\sin(-8t)]. Using the chain rule, we get dydt=3⋅(−8)⋅cos⁡(−8t)=−24cos⁡(−8t)\frac{dy}{dt} = 3 \cdot (-8) \cdot \cos(-8t) = -24\cos(-8t).
  4. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}.dydx=dydtdxdt=−24cos⁡(−8t)20sin⁡(−5t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-24\cos(-8t)}{20\sin(-5t)}.
  5. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}.
    dydx=dydtdxdt=−24cos⁡(−8t)20sin⁡(−5t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-24\cos(-8t)}{20\sin(-5t)}.We can simplify the expression by dividing both the numerator and the denominator by 44.
    dydx=−6cos⁡(−8t)5sin⁡(−5t)\frac{dy}{dx} = \frac{-6\cos(-8t)}{5\sin(-5t)}.

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