Question

A jet of enemy is flying along the path given by y=x^2+7 , Find the nearest distance from where a soldier placed at (3,7) can shoot down the helicopter

Answer 1

nswer
Let (x, y) be any point on y = x^2+ 7
at which helicopter in at a particular moment.
∴ helicopter is at (x, x^2
+ 7 ).
Let d be the distance between the jet at (x, x^2
+ 7 ) and soldier at (3, 7).
∴ d^2
= (x – 3)^2+ (x^2 + 7 – 7)^2
= (x – 3)^2+ (x^2)^2
∴ d^2 = x^4+ x^2 – 6 x + 9
Let f (x) = d^2
= x^4 + x^2 – 6 x + 9
f ' (x) = 4 x^3+ 2 x – 6
= 2 (2 x^3+ x – 3)
= 2 ( x – 1) (2 x^2 + 2 x + 3)
f ' (x) = 0
⇒ 2 (x – 1) (2 x^2+ 2 x + 3) = 0
⇒ x = 1 as we reject imaginary values of x.
f ' ' (x) = 12x^2+ 2
At x = 1, f ' ' (x) = 12 + 2 – 14 > 0 ⇒ f (x) lies a local minimum at x = 1
But x – 1 is only extreme point
∴ f (x) is minimum at x = 1
Nearest distance =d at x=
1=√1+1-6+9=√5

or

Given that an apache helicopter of enemy is flying along the curve given by = x^2 + 7

A soldier placed at (3,7) wants to shoot down the helicopter when it is nearest to him.

Now, the distance of the point from the soldier is S = √[ (x-3)^2 + (y-7)^2 ]

Since the point lies on the curve y = x^2 + 7 ..............1

S = √[ (x-3)^2 + (x^2 + 7 - 7)^2 ]

=> S = √(x^4 + x^2 - 6x + 9)

When the distance is maximum/minimum,

dS/dx = 0

(4x^3 + 2x - 6) / √(x^4 + x^2 - 6x + 9) = 0

=> 4x^3 + 2x - 6 = 0

=> 2x^3 + x -3 = 0

=> (x-1)*(2x^2 + 2x + 3) = 0

The solutions to the above equation are

x = 1, -0.5 ± i*0.5√5

Since the solution cannot have any complex roots.

Hence, x = 1 is the abscissa of the nearest point to the soldier.

From equation 1, we get

y = 12 + 7

=> y = 8

So, the nearest point is (1,8)

Now, nearest distance S = √[ (1-3)^2 + (8-7)^2 ]

=> S = √[ (-2)2 + (1)2 ]

=> S = √[4 + 1]

=> S = √5 units