FPM(2023) January Paper 02 que 4 and solution

Solution

The equation of the curve is:

$$ y = x^3 \sin x $$

Step 1: Differentiate $y$ to find the gradient

The gradient of the tangent at any point is given by $\frac{dy}{dx}$. Using the product rule:

$$ y = u \cdot v, \quad \text{where } u = x^3 \text{ and } v = \sin x $$

The product rule states:

$$ \frac{dy}{dx} = \frac{du}{dx}v + u\frac{dv}{dx} $$

Compute $\frac{du}{dx}$ and $\frac{dv}{dx}$:

$$ \frac{du}{dx} = 3x^2, \quad \frac{dv}{dx} = \cos x $$

Substitute into the product rule:

$$ \frac{dy}{dx} = (3x^2)(\sin x) + (x^3)(\cos x) $$

$$ \frac{dy}{dx} = 3x^2 \sin x + x^3 \cos x $$

Step 2: Find the gradient at $x = \frac{\pi}{2}$

Substitute $x = \frac{\pi}{2}$ into $\frac{dy}{dx}$:

$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 \sin\left(\frac{\pi}{2}\right) + \left(\frac{\pi}{2}\right)^3 \cos\left(\frac{\pi}{2}\right) $$

Simplify:

$$ \sin\left(\frac{\pi}{2}\right) = 1, \quad \cos\left(\frac{\pi}{2}\right) = 0 $$

$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 (1) + \left(\frac{\pi}{2}\right)^3 (0) $$

$$ \frac{dy}{dx} = 3\left(\frac{\pi}{2}\right)^2 = \frac{3\pi^2}{4} $$

Thus, the gradient of the tangent at $x = \frac{\pi}{2}$ is:

$$ m = \frac{3\pi^2}{4} $$

Step 3: Find the point on the curve at $x = \frac{\pi}{2}$

Substitute $x = \frac{\pi}{2}$ into $y = x^3 \sin x$:

$$ y = \left(\frac{\pi}{2}\right)^3 \sin\left(\frac{\pi}{2}\right) $$

$$ y = \left(\frac{\pi}{2}\right)^3 (1) = \frac{\pi^3}{8} $$

The point on the curve is:

$$ \left( \frac{\pi}{2}, \frac{\pi^3}{8} \right) $$

Step 4: Equation of the tangent line

The equation of a straight line is:

$$ y - y_1 = m(x - x_1) $$

Substitute $m = \frac{3\pi^2}{4}$, $x_1 = \frac{\pi}{2}$, and $y_1 = \frac{\pi^3}{8}$:

$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4} \left( x - \frac{\pi}{2} \right) $$

Simplify:

$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4}x - \frac{3\pi^2}{4} \cdot \frac{\pi}{2} $$

$$ y - \frac{\pi^3}{8} = \frac{3\pi^2}{4}x - \frac{3\pi^3}{8} $$

$$ y = \frac{3\pi^2}{4}x - \frac{3\pi^3}{8} + \frac{\pi^3}{8} $$

$$ y = \frac{3\pi^2}{4}x - \frac{2\pi^3}{8} $$

$$ y = \frac{3\pi^2}{4}x - \frac{\pi^3}{4} $$

Final Answer

The equation of the tangent is:

$$ \boxed{ y = \frac{3\pi^2}{4}x - \frac{\pi^3}{4} } $$