Limits, Continuity & Differentiability Question - None

Question

Let $\tan \alpha \cdot x + \sin \alpha \cdot y = \alpha$ and $\alpha \csc \alpha \cdot x + \cos \alpha \cdot y = 1$ be two variable straight lines, $\alpha$ being the parameter. Let P be the point of intersection of the lines. In the limiting position when $\alpha \to 0$, the point P lies on the line:

Accepted Answer

Step 1: Set up the system of equations

We have two lines:

tan α · x + sin α · y = α ... (1)

α csc α · x + cos α · y = 1 ... (2)

Step 2: Solve for the intersection point P(x, y)

From equation (1): tan α · x + sin α · y = α

From equation (2): α · (x/sin α) + cos α · y = 1

Using Cramer's rule with determinant:

D = tan α · cos α - sin α · α/sin α = sin α - α

Step 3: Find x-coordinate using Cramer's rule

D_x = α · cos α - sin α · 1 = α cos α - sin α

x = D_x/D = (α cos α - sin α)/(sin α - α)

x = -(sin α - α cos α)/(α - sin α)

Step 4: Apply limit as α → 0 to find x

As α → 0, both numerator and denominator → 0, so use L'Hôpital's rule:

lim(α→0) (sin α - α cos α)/(α - sin α)

Numerator: d/dα(sin α - α cos α) = cos α - cos α + α sin α = α sin α

Denominator: d/dα(α - sin α) = 1 - cos α

= lim(α→0) (α sin α)/(1 - cos α)

Step 5: Apply L'Hôpital's rule again

Numerator: d/dα(α sin α) = sin α + α cos α

Denominator: d/dα(1 - cos α) = sin α

= lim(α→0) (sin α + α cos α)/sin α = lim(α→0) [1 + α cot α] = 1

Therefore: x = -1 × (1) = wait, recalculating...

Step 5 (Corrected): Recalculate limit

x = -(sin α - α cos α)/(α - sin α) = (α cos α - sin α)/(sin α - α)

Using Taylor: sin α ≈ α - α³/6, cos α ≈ 1 - α²/2

Numerator: α(1 - α²/2) - (α - α³/6) = α - α³/2 - α + α³/6 = -α³/3

Denominator: (α - α³/6) - α = -α³/6

x = (-α³/3)/(-α³/6) = 2

∴ Answer: The limiting position point P lies on the line x = 2, which is option A

Question:

Step-by-Step Solution

Master Limits, Continuity & Differentiability with Mathbee