Differential Calculus-2
Differential Calculus-2
Allen Star Batch
Grade 12
Question:
Let $f: [0, \infty) \to R$ be a continuous, strictly increasing function such that $f^3(x) = \int_0^x uf^2(u)du$. If a normal is drawn to the curve $y = f(x)$ with gradient $-\frac{1}{2}$, then find the intercept made by it on the $y$-axis.
Step-by-Step Solution
Key Concept: Differentiate the integral equation f³(x) = ∫₀ˣ uf²(u)du using Leibniz rule to get 3f²(x)·f'(x) = xf²(x), then use the normal's gradient -1/2 to find that f'(x₁) = 2, which determines the point of tangency where the normal is drawn.
Given $f^3(x) = \int_0^x u^2(t)dt$, differentiate both sides using Leibniz rule: $3f^2(x) \cdot f'(x) = x^2(x)$. This gives $f'(x) = x^2/6$ for $x \geq 0$. The slope of the normal is $-6/x_1$, so the normal equation is $y - 6 = -1(x - 6)$, yielding $y$-intercept of $9$.
Correct Answer: 9