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A curve is defined by the parametric equations x(t)=3cos(-5t) and y(t)=7sin(-8t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=3cos⁡(−5t) x(t)=3 \cos (-5 t) and y(t)=7sin⁡(−8t) y(t)=7 \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)=3cos⁡(−5t) x(t)=3 \cos (-5 t) and y(t)=7sin⁡(−8t) y(t)=7 \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)=3cos⁡(−5t)x(t) = 3\cos(-5t) with respect to tt is dxdt=−3×−5sin⁡(−5t)=15sin⁡(−5t)\frac{dx}{dt} = -3 \times -5\sin(-5t) = 15\sin(-5t).
  3. Divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}: Next, we find dydt\frac{dy}{dt}. The derivative of y(t)=7sin⁡(−8t)y(t) = 7\sin(-8t) with respect to tt is dydt=7⋅−8cos⁡(−8t)=−56cos⁡(−8t)\frac{dy}{dt} = 7 \cdot -8\cos(-8t) = -56\cos(-8t).
  4. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}. So, dydx=−56cos⁡(−8t)15sin⁡(−5t)\frac{dy}{dx} = \frac{-56\cos(-8t)}{15\sin(-5t)}.
  5. Simplify the expression: Now we divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt} to find dydx\frac{dy}{dx}. So, dydx=−56cos⁡(−8t)15sin⁡(−5t)\frac{dy}{dx} = \frac{-56\cos(-8t)}{15\sin(-5t)}.We can simplify the expression by factoring out the negative sign in the numerator and denominator, which gives us dydx=56cos⁡(−8t)−15sin⁡(−5t)\frac{dy}{dx} = \frac{56\cos(-8t)}{-15\sin(-5t)}.

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