Question

If the angle made by the tangent at the point \((x_0, y_0)\) on the curve \(x = 12(t + \sin t \cos t)\), \(y = 12(1 + \sin t)^2\), with \(0<t<\frac{\pi}{2}\), with the positive x-axis is \(\frac{\pi}{3}\), then \(y_0\) is equal to:

• A. \(6(3 + 2\sqrt{2})\)
• B. \(3(7 + 4\sqrt{3})\)
• C. 27
• D. 48

Hint:

To solve problems involving parametric curves and slopes, always recall the derivative formulas and use the trigonometric identities to simplify expressions.

Answer 1

The Correct Option is C

Step 1: Differentiate the given parametric equations with respect to \(t\) to find \(\frac{dy}{dx}\).
We are given: \[ x = 12(t + \sin t \cos t), \quad y = 12(1 + \sin t)^2 \] \[ \frac{dx}{dt} = 12(1 + \cos^2 t - \sin^2 t), \quad \frac{dy}{dt} = 24(1 + \sin t) \cos t \] \[ \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{24(1 + \sin t) \cos t}{12(1 + \cos^2 t - \sin^2 t)} \] Step 2: Use the condition that the angle between the tangent and the positive x-axis is \(\frac{\pi}{3}\). The slope of the tangent is \( \tan \frac{\pi}{3} = \sqrt{3} \). Thus, equate the slope \(\frac{dy}{dx}\) to \(\sqrt{3}\) and solve for \(t\). \[ \frac{24(1 + \sin t) \cos t}{12(1 + \cos^2 t - \sin^2 t)} = \sqrt{3} \] After solving, we find that \( t = \frac{\pi}{6} \). Substituting this value of \( t \) in \( y = 12(1 + \sin t)^2 \), we get: \[ y_0 = 12(1 + \sin \frac{\pi}{6})^2 = 12 \left(1 + \frac{1}{2}\right)^2 = 12 \times \left(\frac{3}{2}\right)^2 = 27 \] Therefore, \( y_0 = 27 \).

Answer 2

Step 1: Understanding the problem
We are given the parametric equations of a curve: \[ x = 12(t + \sin t \cos t), \quad y = 12(1 + \sin t)^2 \] We are asked to find the value of \( y_0 \) when the angle made by the tangent at the point \( (x_0, y_0) \) with the positive x-axis is \( \frac{\pi}{3} \). Step 2: Finding the slope of the tangent
The slope of the tangent to a curve at a point is given by the derivative of \( y \) with respect to \( x \): \[ \text{slope of tangent} = \frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} \] We first compute \( \frac{dx}{dt} \) and \( \frac{dy}{dt} \). 
Step 3: Derivative of \( x \) with respect to \( t \)
\[ x = 12(t + \sin t \cos t) \] Differentiating with respect to \( t \): \[ \frac{dx}{dt} = 12 \left( 1 + \cos^2 t - \sin^2 t \right) = 12 \cdot 2\cos^2 t = 24\cos^2 t \] Step 4: Derivative of \( y \) with respect to \( t \)
\[ y = 12(1 + \sin t)^2 \] Differentiating with respect to \( t \): \[ \frac{dy}{dt} = 12 \cdot 2(1 + \sin t) \cdot \cos t = 24(1 + \sin t) \cos t \] Step 5: Slope of the tangent
The slope of the tangent is: \[ \frac{dy}{dx} = \frac{24(1 + \sin t) \cos t}{24 \cos^2 t} = \frac{1 + \sin t}{\cos t} \] Step 6: Using the given angle with the x-axis
The angle between the tangent and the positive x-axis is \( \frac{\pi}{3} \). The slope of the tangent line is given by the tangent of the angle: \[ \tan\left( \frac{\pi}{3} \right) = \sqrt{3} \] Therefore, we set the slope equal to \( \sqrt{3} \): \[ \frac{1 + \sin t}{\cos t} = \sqrt{3} \] Multiplying both sides by \( \cos t \), we get: \[ 1 + \sin t = \sqrt{3} \cos t \] Squaring both sides: \[ (1 + \sin t)^2 = 3 \cos^2 t \] Using the identity \( \cos^2 t = 1 - \sin^2 t \): \[ (1 + \sin t)^2 = 3(1 - \sin^2 t) \] Expanding both sides: \[ 1 + 2\sin t + \sin^2 t = 3 - 3\sin^2 t \] Simplifying: \[ 4\sin^2 t + 2\sin t - 2 = 0 \] Step 7: Solving the quadratic equation
We solve the quadratic equation: \[ 2\sin^2 t + \sin t - 1 = 0 \] Using the quadratic formula: \[ \sin t = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 2 \cdot (-1)}}{2 \cdot 2} = \frac{-1 \pm \sqrt{1 + 8}}{4} = \frac{-1 \pm 3}{4} \] The two possible solutions are: \[ \sin t = \frac{2}{4} = 0.5 \quad \text{or} \quad \sin t = \frac{-4}{4} = -1 \] Since \( 0 < t < \frac{\pi}{2} \), we take \( \sin t = 0.5 \), which gives: \[ t = \frac{\pi}{6} \] Step 8: Finding \( y_0 \)
Now that we know \( t = \frac{\pi}{6} \), we substitute this value into the equation for \( y \): \[ y = 12(1 + \sin t)^2 = 12\left( 1 + \frac{1}{2} \right)^2 = 12 \times \left( \frac{3}{2} \right)^2 = 12 \times \frac{9}{4} = 27 \] Step 9: Conclusion
Therefore, \( y_0 = 27 \).