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

A curve is defined by the parametric equations x(t)=6sin(9t) x(t)=-6 \sin (-9 t) and y(t)=4cos(4t) y(t)=4 \cos (4 t) . Find dydx \frac{d y}{d x} .\newlineAnswer:

Full solution

Q. A curve is defined by the parametric equations x(t)=6sin(9t) x(t)=-6 \sin (-9 t) and y(t)=4cos(4t) y(t)=4 \cos (4 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)=6sin(9t)x(t) = -6\sin(-9t) with respect to tt is dxdt=6×9×cos(9t)=54cos(9t)\frac{dx}{dt} = -6 \times -9 \times \cos(-9t) = 54\cos(-9t), because the derivative of sin(u)\sin(u) with respect to uu is cos(u)\cos(u) and we apply the chain rule for the derivative of 9t-9t.
  3. Divide to find dydx\frac{dy}{dx}: Next, we find dydt\frac{dy}{dt}. The derivative of y(t)=4cos(4t)y(t) = 4\cos(4t) with respect to tt is dydt=4×4×sin(4t)=16sin(4t)\frac{dy}{dt} = -4 \times 4 \times \sin(4t) = -16\sin(4t), because the derivative of cos(u)\cos(u) with respect to uu is sin(u)-\sin(u) and we apply the chain rule for the derivative of 4t4t.
  4. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}. So, dydx=16sin(4t)54cos(9t)\frac{dy}{dx} = \frac{-16\sin(4t)}{54\cos(-9t)}.
  5. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}. So, dydx=16sin(4t)54cos(9t)\frac{dy}{dx} = \frac{-16\sin(4t)}{54\cos(-9t)}.We can simplify the expression by dividing both the numerator and the denominator by 22, which gives us dydx=8sin(4t)27cos(9t)\frac{dy}{dx} = \frac{-8\sin(4t)}{27\cos(-9t)}.

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