Q. 56 - Calculus

Question

An airplane leaves Chicago at noon and travels south at $500$ miles per hour. Another airplane is traveling east towards Chicago at $650$ miles per hour and arrives at $2:00$ p.m. When were these two airplanes closest to each other, and how far apart were they at that time?

Step-by-Step Solution

Answer

Ans:  The distance between the two airplanes at time $t=\raisebox{1ex}{$338$}\!\left/ \!\raisebox{-1ex}{$269$}\right.$ was approximately $793$ miles.

At $t=\raisebox{1ex}{$338$}\!\left/ \!\raisebox{-1ex}{$269$}\right.$, the two airplanes were closest to each other, and the distance between the two planes at that time was $793$ miles.

Step 1. Given information.

given,

     An airplane leaves Chicago at noon and travels south at $500$ miles per hour. Another airplane is traveling east towards Chicago at $650$ miles per hour and arrives at $2:00$ p.m.

 

Step 2. The objective is to determine the time when the two airplanes were closest to each other and the distance between the two planes at that time.

Consider that at times $t$ the two planes were closest to each other.

Then the distance of the plane traveling away and towards Chicago at time $t$ is $500t$ and $1300-650t$.

Draw a triangle similar to the one given in the question and then label the distances.

Step 3. The Pythagorean theorem states that, in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the side opposite and side adjacent, i.e., c 2 = a 2 + b 2 a , b and c are the side opposite, side adjacent, and hypotenuse respectively.

By the Pythagorean theorem, the distance between the two planes at time $t$ is

  $D^2=(1300-650t{)}^2+(500t{)}^2.$

Consider  $D^2=f\left(t\right)\text{, then }f\left(t\right)=(1300-650t{)}^2+(500t{)}^2$

Calculate the minimum value of $f\left(t\right)=(1300-650t{)}^2+(500t{)}^2$ to obtain the time when the planes were closest to each other.

Simplify the function $f\left(t\right)=(1300-650t{)}^2+(500t{)}^2$ by squaring the binomial.

    $\begin{array}{r}f\left(t\right)=(1300-650t{)}^2+(500t{)}^2\\ =422500t^2-1690000t+1690000+250000\\ =672500t^2-1690000t+1690000\end{array}$

Step 4. The chain rule for differentiation states that d y d x = d y d t ⋅ d t d x ,  where  y = f ( t ) is a differentiable function of t and t = g ( x ) ) is a differentiable function of x .

Differentiate $f\left(t\right)=672500t^2-1690000t+1690000$ with respect to $t$ the use of chain rule to obtain the derivative.

  $\begin{array}{r}{f}^{\mathrm{\prime }}\left(t\right)=\frac{d}{dt}\left(672500t^2-1690000t+1690000\right)\\ =1345000t-1690000\end{array}$

Step 5. The critical point of a function f is a point in the domain of the function where the function is either not differentiable or the derivative is equal to zero.

Set up the equation $1345000t-1690000=0$ and then solve for $t$ to obtain the critical point of $D$.

    $\begin{array}{r}1345000t-1690000=0\\ 1345000t=1690000\\ t=\frac{1690000}{1345000}\\ t=\frac{338}{269}\end{array}$

Step 6. Differentiate the function f ′ ( t ) = 1345000 t − 1690000 with respect to t to obtain the second derivative.

   $\begin{array}{r}{D}^{\prime \prime }\left(t\right)=\frac{d}{dt}(1345000t-1690000)\\ =\frac{d}{dt}\left(1345000t\right)-\frac{d}{dt}\left(1690000\right)\\ =1345000\end{array}$

Step 7. The second-derivative test states that if x = c is the location of a critical point of a function f and if both f and f ' are differentiable and f ' ' is continuous on an interval around x = c , then f has a (i) local maximum at x = c , if f ' ' ( c ) is negative, (ii) local minimum at x = c , if f ' ' ( c ) (c) is positive, (iii) the test is inconclusive if f ' ' ( c ) = 0 .

Since the second derivative is positive, it confirms that at $t=\raisebox{1ex}{$338$}\!\left/ \!\raisebox{-1ex}{$269$}\right.$, the distance between the two airplanes was the least.

Substitute $t=\frac{338}{269}\text{ in }{D}^2=(1300-650t{)}^2+(500t{)}^2$ and then solve for $D$.

    $\begin{array}{r}{D}^2={\left[1300-650\left(\frac{338}{269}\right)\right]}^2+{\left[500\left(\frac{338}{269}\right)\right]}^2\\ D^2\approx 628253\\ D\approx \sqrt{628253}\\ D\approx 793\end{array}$

The distance between two airplanes at the time $t=\frac{338}{269}$ was approximately $793$ miles.

At, $t=\frac{338}{269}$, the two airplanes were closest to each other, and the distance between the two planes at that time was $793$ miles.