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A curve is defined by the parametric equations x(t)=-8e^(-9t) and y(t)=sin(-5t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=−8e−9t x(t)=-8 e^{-9 t} and y(t)=sin⁡(−5t) y(t)=\sin (-5 t) . Find dydx \frac{d y}{d x} .\newlineAnswer:

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Q. A curve is defined by the parametric equations x(t)=−8e−9t x(t)=-8 e^{-9 t} and y(t)=sin⁡(−5t) y(t)=\sin (-5 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 first need to find dxdt\frac{dx}{dt} and dydt\frac{dy}{dt} separately.\newlineFor x(t)=−8e−9tx(t) = -8e^{-9t}, we differentiate with respect to tt to find dxdt\frac{dx}{dt}.\newlinedxdt=ddt[−8e−9t]\frac{dx}{dt} = \frac{d}{dt} [-8e^{-9t}]\newlineUsing the chain rule, we get dxdt=−8⋅(−9)⋅e−9t\frac{dx}{dt} = -8 \cdot (-9) \cdot e^{-9t}\newlinedxdt=72e−9t\frac{dx}{dt} = 72e^{-9t}
  2. Find dydt\frac{dy}{dt}: Now, we differentiate y(t)=sin⁡(−5t)y(t) = \sin(-5t) with respect to tt to find dydt\frac{dy}{dt}.
    dydt=ddt[sin⁡(−5t)]\frac{dy}{dt} = \frac{d}{dt} [\sin(-5t)]
    Using the chain rule, we get dydt=cos⁡(−5t)⋅(−5)\frac{dy}{dt} = \cos(-5t) \cdot (-5)
    dydt=−5cos⁡(−5t)\frac{dy}{dt} = -5\cos(-5t)
    Since cosine is an even function, cos⁡(−5t)=cos⁡(5t)\cos(-5t) = \cos(5t), so we can simplify this to:
    dydt=−5cos⁡(5t)\frac{dy}{dt} = -5\cos(5t)
  3. Calculate dydx\frac{dy}{dx}: To find dydx\frac{dy}{dx}, we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}.
    dydx=dydtdxdt\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}
    Substitute the values we found for dydt\frac{dy}{dt} and dxdt\frac{dx}{dt}.
    dydx=−5cos⁡(5t)72e−9t\frac{dy}{dx} = \frac{-5\cos(5t)}{72e^{-9t}}
  4. Simplify dydx\frac{dy}{dx}: We can simplify the expression for dydx\frac{dy}{dx} by dividing both the numerator and the denominator by the common factor if there is any. However, in this case, there is no common factor, so the expression is already in its simplest form.

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