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A curve is defined by the parametric equations x(t)=-9cos(4t) and y(t)=-6sin(5t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=−9cos⁡(4t) x(t)=-9 \cos (4 t) and y(t)=−6sin⁡(5t) y(t)=-6 \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)=−9cos⁡(4t) x(t)=-9 \cos (4 t) and y(t)=−6sin⁡(5t) y(t)=-6 \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 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)=−9cos⁡(4t)x(t) = -9\cos(4t) with respect to tt is dxdt=ddt[−9cos⁡(4t)]\frac{dx}{dt} = \frac{d}{dt} [-9\cos(4t)]. Using the chain rule, dxdt=36sin⁡(4t)\frac{dx}{dt} = 36\sin(4t).
  3. Divide for dydx\frac{dy}{dx}: Next, we find dydt\frac{dy}{dt}. The derivative of y(t)=−6sin⁡(5t)y(t) = -6\sin(5t) with respect to tt is dydt=ddt[−6sin⁡(5t)]\frac{dy}{dt} = \frac{d}{dt} [-6\sin(5t)].\newlineUsing the chain rule, dydt=−30cos⁡(5t)\frac{dy}{dt} = -30\cos(5t).
  4. Simplify dydx\frac{dy}{dx}: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}.dydx=dydtdxdt=−30cos⁡(5t)36sin⁡(4t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-30\cos(5t)}{36\sin(4t)}.
  5. Simplify dydx\frac{dy}{dx}: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}.
    dydx=dydtdxdt=−30cos⁡(5t)36sin⁡(4t)\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{-30\cos(5t)}{36\sin(4t)}.We simplify the expression for dydx\frac{dy}{dx}.
    dydx=−3036⋅cos⁡(5t)sin⁡(4t)=−56⋅cos⁡(5t)sin⁡(4t)\frac{dy}{dx} = \frac{-30}{36} \cdot \frac{\cos(5t)}{\sin(4t)} = \frac{-5}{6} \cdot \frac{\cos(5t)}{\sin(4t)}.

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