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Question: An apache helicopter of enemy is flying along the curve given by \(y = {x^2} + 7\). A soldier placed...

An apache helicopter of enemy is flying along the curve given by y=x2+7y = {x^2} + 7. A soldier placed at (3,7)(3,7), wants to shoot down the helicopter when nearest to him. Find the nearest distance.

Explanation

Solution

In the given question we are given that helicopter of the evening is flying along the curve and we have to find the points where the soldier can shoot down the helicopter when it is nearest to him so for that we will assume a point nearest to the soldier and assume as coordinate as (xx,yy) and find the distance between the nearest point and the soldier by using the distance formula.
S=(x2−x1)2+(y2−y1)2\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}
Here S is the distance and x2{x_2} and x1{x_1} be the xx coordinate and y2{y_2} y1{y_1} be the yy coordinates as the point lies on the curve we will substitute the value and they find the minimum |maximum distance by finding the solution of the equation and find the coordinates of xx and yy and then finding the value of distance.

Complete step by step answer:
Step1: The given is-y=x2+7y = {x^2} + 7
Let s be the distance between helicopter soldier at (3,7)(3,7)
S=(x2−x1)2+(y2−y1)2\sqrt {{{({x_2} - {x_1})}^2} + {{({y_2} - {y_1})}^2}}
S=(x−3)2+(y−7)2\sqrt {{{(x - 3)}^2} + {{(y - 7)}^2}} …..(1)
Since the point lies on the curve hencey=x2+7y = {x^2} + 7substituting the value ofyyin the equation (1) we get
⇒S=(x−3)2+((x2+7)−7)2\Rightarrow S=\sqrt {{{(x - 3)}^2} + {{(({x^2} + 7) - 7)}^2}}
⇒S=(x−3)2+(x)4\Rightarrow S=\sqrt {{{(x - 3)}^2} + {{(x)}^4}}
Step2: We need to find nearest distance i.e. minimum value of S
Let f(X)=S2f(X) = S^2
⇒f(X)=(x−3)2+x4\Rightarrow f(X)={(x - 3)^2} + {x^4}
when f(X) is minimum, S is minimum
for maxima and minima we will find f’(X)
f’(X)=2(x−3)+4x3 = 2(x - 3) + 4{x^3}
=2x−6+4x3= 2x - 6 + 4{x^3}
On further solving
F’(X)=4x3−2x−6 = 4{x^3} - 2x - 6
Step3: Factorizing f’(X)
Thus,
F’(X)=(x−1)(4x3+4x+6) = (x - 1)(4{x^3} + 4x + 6)
For maxima and minima f’(X)=0 = 0
Hence (x−1)=0(x - 1) = 0
x=1x = 1
⇒2x2+2x+3=0\Rightarrow 2{x^2} + 2x + 3 = 0
Using the quadratic formula:
⇒x=−2±4−4(2)(3)4\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 4(2)(3)} }}{4}
⇒x=−2±4−244\Rightarrow x = \dfrac{{ - 2 \pm \sqrt {4 - 24} }}{4}
⇒x=−2±−204\Rightarrow x = \dfrac{{ - 2 \pm \sqrt { - 20} }}{4}
This is not possible as there are no real roots
Hence there is only one point,
⇒x=1\Rightarrow x = 1
This is either maxima or minima
Hence we find f’’(X)
Step4: f’(X)=(4x3+2x−6)′ = (4{x^3} + 2x - 6)'
F’’(X)=12x2+2 = 12{x^2} + 2
Finding value atx=1x = 1,
⇒f′′(1)=12(1)2+2\Rightarrow f''(1) = 12{(1)^2} + 2
⇒f′′(1)=12+2\Rightarrow f''(1) = 12 + 2
⇒f′′(1)=14\Rightarrow f''(1) = 14
Since f’’(X)>0 > 0
∴x=1\therefore x = 1is the minima
The value of f(1)f(1)is
⇒f(1)=(1−3)2+14\Rightarrow f(1) = {(1 - 3)^2} + {1^4}
⇒f(1)=4+1\Rightarrow f(1) = 4 + 1
=5= 5
Hence minimum distance between soldier and helicopter
⇒S=f(1)\Rightarrow S = \sqrt {f(1)}
⇒S=5\Rightarrow S = \sqrt 5

Hence the minimum distance is 5\sqrt 5 .

Note:
In such type of question students mainly get confused in finding the nearest distance they don’t go towards the concept of finding the minimum and maximum value so they should be kept in mind to find the minimum or nearest distance or maximum value they use the concept of maxima and minima by finding derivatives they should also water the calculation