Question

Two cyclists leave simultaneously from the Math Department at UT. One travels north at 16 mph, the other travels east at 30 mph. Determine the rate at which the distance between them is changing after 2 hours of riding.

Answer 1

Answer:

The rate at which the distance between them is changing after 2 hours of riding is 34 mph.

Step-by-step explanation:

Let $x_1\ and\ x_2$ be the distances traveled in North and East directions respectively at any time 't'.

Given:

Rate of change of distance for cyclist traveling North is, $\frac{dx_1}{dt}=16\ mph$

Rate of change of distance for cyclist traveling East is, $\frac{dx_2}{dt}=30\ mph$

Time (t) = 2 hours

Now, the given scenario can be represented using a right angled triangle ABC shown below.

From the triangle, AB is the distance traveled North and BC is distance traveled East. AC is the distance between the two cyclists.

Let AC = 'x' at any time 't'.

Using Pythagoras theorem, we have:

$AC^2=AB^2+BC^2\\\\x^2=x_1^2+x_2^2------(1)$

Differentiating both sides with respect to time 't', we get:

$2x\frac{dx}{dt}=2x_1\frac{dx_1}{dt}+2x_2\frac{dx_2}{dt}\\\\x\frac{dx}{dt}=16x_1+30x_2---------(2)$

Now, we are asked to find the rate of change of distance between the cyclists which is nothing but the derivative of 'x' with time 't'.

So, in order to find that, we first find $x_1,x_2,\ and\ x$ when time $t = 2$

We know that, distance = speed × time

So, $x_1=16\times 2=32\ mi$ and $x_2=30\times 2=60\ mi$

Therefore, from equation (1), the value of 'x' is given as:

$x=\sqrt{32^2+60^2}=68\ mi$

Now, plug in the values of $x_1,x_2,\ and\ x$ in equation (2). This gives,

$68\times \frac{dx}{dt}=32\times 16+60\times 30\\\\68\times \frac{dx}{dt}=512+1800\\\\\frac{dx}{dt}=\frac{2312}{68}=34\ mph$

Therefore, the rate at which the distance between them is changing after 2 hours of riding is 34 mph.