Question

An Apache helicopter of an 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.

Answer 1

Hint: This problem demands the use of derivatives. Here, at first we will find any general point on this curve and then we will calculate its distance from the given point using distance formula. In order to get the minimum distance, we will make the derivative of the distance equal to zero.

Complete step-by-step answer:
Since, the curve given to us is:
$y=x^2+7...........\left(1\right)$
So, any general point on this given curve will be of the form $(x,x^2+7)$ .
To calculate the distance of the point (3, 7) from this point can be calculated using the distance formula.
We know that the distance formula is given as:
$d=\sqrt{{(x_2-x_1)}^2+{(y_2-y_1)}^2}$

So, using this formula we get the distance between the general point and (3, 7) as:
$\begin{array}{rl}& d=\sqrt{{(3-x)}^2+{\{7-(x^2+7)\}}^2}\\ & d=\sqrt{{(3-x)}^2+{(7-x^2-7)}^2}\\ & d=\sqrt{{(3-x)}^2+x^4}\end{array}$

Since, we have to find the minimum value of d. At the point of minimum the derivative of d must be zero.
So, we will differentiate d which is a function of x with respect to x and equate it to zero to get the point of minimum.
$\begin{array}{rl}& {\displaystyle \frac{d\left(d\right)}{dx}}=0\\ & {\displaystyle \frac{1}{2\sqrt{{(3-x)}^2+x^4}}}\times \{2(3-x)(-1)+4x^3\}=0\\ & \{-2(3-x)+4x^3\}=0\\ & -6+2x+4x^3=0\end{array}$

If we substitute x = 1 in this equation, we get:
$-6+2\times 1+4{\left(1\right)}^3=-6+2+4=0$
It means that x = 1 is a solution of this equation and hence, (x-1) is a factor of $-6+2x+4x^3$ .
So, we have:
$(x-1)(4x^2+4x+6)=0$

Now the discriminant of the quadratic equation $(4x^2+4x+6)$ is = $4^2-4\times 4\times 6=16-84=-48$.
Since, the discriminant is zero so, this quadratic equation does not have any real roots.
Now we will double differentiate the d and check whether it gives a positive value at x=1.

If it gives a positive value at x =1 , then x = 1 will be a point of minimum.
${\displaystyle \frac{d\left(d^{\prime }\right)}{dx}}={\displaystyle \frac{\{{\displaystyle \frac{1}{2\sqrt{{(3-x)}^2+x^4}}}\times (-2)(-1)+12x^3\}-\{(-2)\times (3-x)+4x^3\}\times {\displaystyle \frac{(-1)}{4\times {\{{(3-x)}^2+x^4\}}^{{\displaystyle \frac{3}{2}}}}}}{4\{{(3-x)}^2+x^4\}}}$
On putting x=1, we get:
$\begin{array}{rl}& {\displaystyle \frac{d\left(d^{\prime }\right)}{dx}}={\displaystyle \frac{\{{\displaystyle \frac{1}{2\sqrt{4+1}}}\times (2+12)\}-\{(-4+4)\times {\displaystyle \frac{(-1)}{4\times {(4+1)}^{{\displaystyle \frac{3}{2}}}}}\}}{4(4+1)}}\\ & ={\displaystyle \frac{{\displaystyle \frac{14}{2\sqrt{5}}}-0}{20}}\\ & ={\displaystyle \frac{14}{40\sqrt{5}}}\end{array}$
So, it comes out to be greater than zero at x=1. Therefore, x=1 is the minimum.
On substituting x = 1 in equation (1), we get:
$y={\left(1\right)}^2+7=8$

So, nearest distance is the distance between (3, 7) and (1, 8) and using the distance formula this distance will be:
$\begin{array}{rl}& =\sqrt{{(3-1)}^2+{(7-8)}^2}\\ & =\sqrt{4+1}\\ & =\sqrt{5}\\ & =2.23\end{array}$
Hence, the nearest distance is 2.23 units.

Note: Students should note that at the point of minimum the derivative of a function is zero. Also, at the point of minimum the double derivative of the function is positive.