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A curve is defined by the parametric equations x(t)=9cos(8t) and y(t)=-3cos(-t). Find (dy)/(dx). Answer:

A curve is defined by the parametric equations x(t)=9cos⁡(8t) x(t)=9 \cos (8 t) and y(t)=−3cos⁡(−t) y(t)=-3 \cos (-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⁡(8t) x(t)=9 \cos (8 t) and y(t)=−3cos⁡(−t) y(t)=-3 \cos (-t) . Find dydx \frac{d y}{d x} .\newlineAnswer:
  1. Find dxdt\frac{dx}{dt}: To find the derivative of yy with respect to xx (dydx\frac{dy}{dx}) for parametric equations, we need to find dydt\frac{dy}{dt} and dxdt\frac{dx}{dt} and then divide dydt\frac{dy}{dt} by dxdt\frac{dx}{dt}. First, let's find dxdt\frac{dx}{dt}. dxdt=ddt[9cos⁡(8t)]\frac{dx}{dt} = \frac{d}{dt} [9\cos(8t)] Using the chain rule, the derivative of yy00 with respect to yy11 is yy22 multiplied by the derivative of yy33 with respect to yy11, which is yy55. So, yy66
  2. Find dydt\frac{dy}{dt}: Now, let's find dydt\frac{dy}{dt}.
    dydt=ddt[−3cos⁡(−t)]\frac{dy}{dt} = \frac{d}{dt} [-3\cos(-t)]
    Using the chain rule, the derivative of cos⁡(−t)\cos(-t) with respect to tt is sin⁡(−t)\sin(-t) multiplied by the derivative of −t-t with respect to tt, which is −1-1.
    So, dydt=3sin⁡(−t)\frac{dy}{dt} = 3\sin(-t)
    Since dydt\frac{dy}{dt}00, we can simplify this to dydt\frac{dy}{dt}11
  3. Calculate dydx\frac{dy}{dx}: Now that we have both derivatives, we can find dydx\frac{dy}{dx} by dividing 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=−3sin⁡(t)−72sin⁡(8t)\frac{dy}{dx} = \frac{-3\sin(t)}{-72\sin(8t)}
  4. Simplify expression: We can simplify the expression by dividing both the numerator and the denominator by −3sin⁡(t)-3\sin(t).(dydx)=sin⁡(t)24sin⁡(8t)(\frac{dy}{dx}) = \frac{\sin(t)}{24\sin(8t)}However, this simplification assumes that sin⁡(t)\sin(t) is not zero, as division by zero is undefined. Since sin⁡(t)\sin(t) can be zero for some values of tt, we must be careful not to cancel out terms that could lead to division by zero. In this case, we can proceed with the simplification because we are looking for a general expression for (dydx)(\frac{dy}{dx}), not its value at specific points.
  5. Final derivative: The final step is to simplify the expression further if possible. However, without additional information about the relationship between sin⁡(t)\sin(t) and sin⁡(8t)\sin(8t), we cannot simplify the expression any further.\newlineTherefore, the derivative of yy with respect to xx, given the parametric equations, is:\newline(dydx)=sin⁡(t)24sin⁡(8t)(\frac{dy}{dx}) = \frac{\sin(t)}{24\sin(8t)}

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