Question

A particle is moving along the curve \( y = 8x + \cos y \), where \( 0 \leq y \leq \pi \). If at a point the ordinate is changing 4 times as fast as the abscissa, then the coordinates of the point are:

• A. \(\left(\frac{\pi}{16}, \frac{\pi}{2}\right)\)
• B. \(\left(-\frac{1}{8}, 0\right)\)
• C. \(\left(\frac{1}{8}, 0\right)\)
• D. \(\left(-\frac{\pi}{2}, -\frac{\pi}{16}\right)\)
• E. \(\left(\frac{\pi}{2}, \frac{9\pi}{16}\right)\)

Hint:

When dealing with implicit differentiation and equations involving trigonometric functions, always consider the specific domain and range values applicable to the function and the physical context of the problem.

Answer 1

The Correct Option is A

Differentiate implicitly with respect to \(x\): \[ \frac{dy}{dx} = 8 - \sin y \frac{dy}{dx} \] \[ \frac{dy}{dx} + \sin y \frac{dy}{dx} = 8 \] \[ \frac{dy}{dx}(1 + \sin y) = 8 \] \[ \frac{dy}{dx} = \frac{8}{1 + \sin y} \] Given that the ordinate (\(y\)) is changing four times as fast as the abscissa (\(x\)): \[ \frac{dy}{dx} = 4 \] \[ 4 = \frac{8}{1 + \sin y} \] \[ 1 + \sin y = 2 \] \[ \sin y = 1 \] The \(y\)-value that satisfies \(\sin y = 1\) within the given range is: \[ y = \frac{\pi}{2} \] Substitute \(y = \frac{\pi}{2}\) back into the original equation to find \(x\): \[ y = 8x + \cos \left(\frac{\pi}{2}\right) \] \[ \frac{\pi}{2} = 8x + 0 \] \[ x = \frac{\pi}{16} \] Thus, the coordinates of the point are \(\left(\frac{\pi}{16}, \frac{\pi}{2}\right)\).