Question

Two boats leave the port at the same time with one boat travelling north at 15 knots per hour and the other boat travelling west at 12 knots per hour. How fast is the distance between the boats changing after 2 hours?

Answer 1

Hint: In order to solve this question, first we will apply the Pythagoras theorem and we will find out the expression for the distance between the two boats. Then we will do differentiation and further we will put the value of time as 2. Then on putting the value of distance and time in the differential equation, we get the final answer.

Complete step-by-step answer:
In this particular question, let us assume the distance between the two boats be \[x\] and the number of hours they have travelled be $h$.
We know that,
$speed = \dfrac{{dis\tan ce}}{{time}}$
$dis\tan ce = speed \times time$
So, the distance travelled by the first boat is ${x_1} = 15h$
Similarly, the distance travelled by the second boat is ${x_2} = 12h$
On applying Pythagoras theorem, we get,
${x^2} = x_1^2 + x_2^2$
${x^2} = {(15h)^2} + {(12h)^2}$
On further solving, we get,
${x^2} = 225{h^2} + 144{h^2}$
${x^2} = 369{h^2}$
On differentiating the above equation with respect to time, we get,
$2x\left( {\dfrac{{dx}}{{dt}}} \right) = 738h......(1)$
Now, we need to find out the distance which is present between the two boats when they have travelled for two hours.
Distance travelled by the first boat after two hours $ = 15 \times 2 = 30knots$
Distance travelled by the second boat after two hours $ = 12 \times 2 = 24knots$
Now, on applying Pythagoras theorem on this, we get,
${x^2} = {(30)^2} + {(24)^2}$
${x^2} = 1476$
On taking square root on both the sides, we get,
$x = \sqrt {1476} $
Now, on putting $x = \sqrt {1476} $ and $h = 2$ in equation (1), we get,
$2(\sqrt {1476} )\left( {\dfrac{{dx}}{{dt}}} \right) = 738(2)$
On cancelling the common terms on both the sides of the above equation, we get,
$\sqrt {1476} \left( {\dfrac{{dx}}{{dt}}} \right) = 738$
$\left( {\dfrac{{dx}}{{dt}}} \right) = \dfrac{{738}}{{\sqrt {1476} }}$
On further solving the above equation, we get,
$\left( {\dfrac{{dx}}{{dt}}} \right) = \dfrac{{\sqrt {1476} }}{2}$
So, the value $\left( {\dfrac{{dx}}{{dt}}} \right) = \dfrac{{\sqrt {1476} }}{2}$ represents the speed between the boats which is changing after 2 hours.

Note: It is important to note that the final answer that we have obtained in this question is in terms of knot per second. To convert it into metres per second, we will divide the above expression with $1.944$ and get the answer in metres per second.