Q. 265.0( 1 Vote )

# A jet of enemy is

Given: The curve is y = x2 + 2

To find: minimum distance between the soldier and the jet

Proof: Let jet be at point (x, y) = (x, x2 + 2)

Let d be the distance between jet and soldier at (3, 2)

We need to find the minimum distance .i.e. minimum value of d

Let f(x) = d2 [to make calculation easy]

f(x) = (x – 3)2 + x4

When f(x) is minimum, d is minimum

Finding f’(x)

f’(x) = 2 × (x – 3) × 1 + 4 × x3

= 2x – 6 + 4x3

= 4x3 + 2x – 6

Factorizing f’(x)

f’(1) = 4(1)3 + 2(1) – 6

= 4 + 2 – 6

= 6 – 6

= 0

Hence, (x – 1) is a factor of 4x3 + 2x – 6

Thus,

f’(x) = (x – 1)(4x2 + 4x + 6)

Hence, f’(x) = 0 gives

either x – 1 = 0

x = 1

or 4x2 + 4x + 6 = 0

2(2x2 + 2x +3) = 0

2x2 + 2x + 3 = 0

Now, we find the value of x

Here, a = 2, b = 2 and c = 3

This is not possible as there are no real roots

Hence, there is only one point x = 1

This is either maxima or minima

Hence, we find f’’(x)

f’’(x) = 3 × 4x2 + 2

f’’(x) = 12x2 + 2

Finding value at x = 1

f’’(1) = 12(1)2 + 2

= 12 + 2

= 14 > 0

f’’(x) > 0

x = 1 is the minima

The value of f(1) is

f(1) = (1 – 3)2 + (1)4

= (-2)2 + 1

= 4 + 1

= 5

Hence, minimum distance between soldier and jet

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