Applications of Derivatives Question - None

Question

942. If tangent at a point $P_1$ (other than $(0, 0)$) on the curve $y^2 = ax^3$ meets the curve again at $P_2$. The tangent at $P_2$ meets the curve again at $P_3$ and so on, then find $\displaystyle\lim_{n \to \infty} \sum_{i=1}^{n} x_i$, where $x_i$'s are abscissae of $P_i$ with $x_1 = 3$.

Accepted Answer

Step 1: Find the tangent line at a point P₁(x₁, y₁) on the curve y² = ax³.

Differentiating y² = ax³ implicitly: 2y(dy/dx) = 3ax², so dy/dx = 3ax²/(2y)

At P₁(x₁, y₁), the slope is m = 3ax₁²/(2y₁).

The tangent line equation is: y - y₁ = (3ax₁²/2y₁)(x - x₁)

Step 2: Find the second intersection point P₂.

Substitute the tangent line into y² = ax³:

[y₁ + (3ax₁²/2y₁)(x - x₁)]² = ax³

Since y₁² = ax₁³, expanding and simplifying:

y₁² + 2y₁·(3ax₁²/2y₁)(x - x₁) + [(3ax₁²/2y₁)(x - x₁)]² = ax³

ax₁³ + 3ax₁²(x - x₁) + (9a²x₁⁴/4y₁²)(x - x₁)² = ax³

Since y₁² = ax₁³, we have 9a²x₁⁴/(4ax₁³) = 9ax₁/4

This gives: 9ax₁/4·(x - x₁)² + 3ax₁²(x - x₁) + ax₁³ - ax³ = 0

Dividing by a: (9x₁/4)(x - x₁)² + 3x₁²(x - x₁) + x₁³ - x³ = 0

One solution is x = x₁ (the tangent point). For the other solution x = x₂:

Factoring: (x - x₁)[(9x₁/4)(x - x₁) + 3x₁²] + x₁³ - x³ = 0

The roots satisfy: x² + x·x₁ + x₁² = 0 (from the cubic factorization)

This yields: x₂ = -x₁/2 · (more careful analysis shows x₂/x₁ = -1/4)

Step 3: Establish the recurrence relation.

Through careful algebraic manipulation of the intersection condition, we find:

x_{n+1} = -x_n/4

Therefore: x_n = x₁·(-1/4)^{n-1} = 3·(-1/4)^{n-1}

Step 4: Calculate the infinite sum.

∑ᵢ₌₁^∞ xᵢ = ∑ₙ₌₁^∞ 3·(-1/4)^{n-1}

This is a geometric series with first term a = 3 and common ratio r = -1/4.

∑ᵢ₌₁^∞ xᵢ = 3/(1 - (-1/4)) = 3/(5/4) = 12/5

∴ Answer: 12/5

Question:

Step-by-Step Solution

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