A helicopter of the enemy is flying along the curve given by  y=x^2+7. A soldier, placed at (3, 7), wants to shoot down the helicopter when it is nearest to him. Find the nearest distance. 

 

 

 

 
 
 
 
 

Answers (1)

Curve  y=x^2+7 and  P(x,x^2+7) is cordinate and  Q(3,7).

Distance  PQ=\sqrt{(x-3)^2+(x^2-7+7)^2}

(PQ)^2=(x-3)^2+x^4

Differentiate wrt x,

f(x)=\frac{\mathrm{d} }{\mathrm{d} x}(PQ)^2=2(x-3)+4x^3=0

2x-6+4x^3=0

x-3+2x^3=0

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

In the quadratic part, D<0. Therefore no roots.

\therefore x=1  is the root of the equation.

f'(1)<0

Minima is at x = 1

PQ=\sqrt{(x-3)^2+(x^2)^2}

At x=1, 

PQ=\sqrt{4+1}=\sqrt{5}

Therefore, the minimum distance is  \sqrt{5}.

 

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