Applications of Derivatives Question - None

Question

Ex. 24(D): If the area of a triangle formed by the normal at the point (1, 0) on the curve x = e^{\sin y} with the axes is \frac{|2t+1|}{6} sq units, then the value of t is

Accepted Answer

Solution:

Given: x = e^{\sin y}

Differentiating: \frac{dx}{dy} = \cos y \cdot e^{\sin y}

At point (1, 0): \frac{dx}{dy} = \cos(0) \cdot e^0 = 1

Therefore: \frac{dy}{dx} = 1

Slope of normal at (1, 0): m = -\frac{1}{1} = -1

Equation of normal: y - 0 = -1(x - 1) \Rightarrow x + y = 1

X-intercept: (1, 0)\, Y-intercept: (0, 1)

Area of triangle with axes: ext{Area} = \frac{1}{2} imes 1 imes 1 = \frac{1}{2}

Given: \frac{|2t+1|}{6} = \frac{1}{2}

|2t+1| = 3 \Rightarrow 2t + 1 = \pm 3

t = 1 ext{ or } t = -2

∴ Answer: (p, s)

Question:

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