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A curve is defined by the parametric equations x(t)=-2sin(-9t) and y(t)=-7t^(3)+5t^(2)+10 t. Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=−2sin⁡(−9t) x(t)=-2 \sin (-9 t) and y(t)=−7t3+5t2+10t y(t)=-7 t^{3}+5 t^{2}+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)=−2sin⁡(−9t) x(t)=-2 \sin (-9 t) and y(t)=−7t3+5t2+10t y(t)=-7 t^{3}+5 t^{2}+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, let's find dxdt\frac{dx}{dt}. The derivative of x(t)=−2sin⁡(−9t)x(t)=-2\sin(-9t) with respect to tt is found using the chain rule.\newlinedxdt=ddt[−2sin⁡(−9t)]=−2⋅cos⁡(−9t)⋅(−9)=18cos⁡(−9t)\frac{dx}{dt} = \frac{d}{dt} [-2\sin(-9t)] = -2 \cdot \cos(-9t) \cdot (-9) = 18\cos(-9t)
  3. Calculate dydx\frac{dy}{dx}: Now, let's find dydt\frac{dy}{dt}. The derivative of y(t)=−7t3+5t2+10ty(t)=-7t^3+5t^2+10t with respect to tt is found by differentiating each term separately.dydt=ddt[−7t3+5t2+10t]=−21t2+10t+10\frac{dy}{dt} = \frac{d}{dt} [-7t^3 + 5t^2 + 10t] = -21t^2 + 10t + 10
  4. Calculate dydx\frac{dy}{dx}: Now, let's find dydt\frac{dy}{dt}. The derivative of y(t)=−7t3+5t2+10ty(t)=-7t^3+5t^2+10t with respect to tt is found by differentiating each term separately.dydt=ddt[−7t3+5t2+10t]=−21t2+10t+10\frac{dy}{dt} = \frac{d}{dt} [-7t^3 + 5t^2 + 10t] = -21t^2 + 10t + 10Finally, we find dydx\frac{dy}{dx} by dividing dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}.dydx=dydt/dxdt=−21t2+10t+1018cos⁡(−9t)\frac{dy}{dx} = \frac{dy}{dt} / \frac{dx}{dt} = \frac{-21t^2 + 10t + 10}{18\cos(-9t)}

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