Question

You and a friend both leave the same restaurant to drive home. You are heading directly west at 30 miles per hour and he or she is heading directly south at 40 miles per hour. After half an hour, how fast (in mph) is the distance between you changing?

Answer 1

Answer:

$\frac{dAB}{dt} = = 50 miles/ hour$

Explanation:

let A represent me and B represent my friend

A speed$\frac{dOA}{dt}$=30 m/hr toward west

B speed$\frac{dOB}{dt}$ =40 m/hr toward south

after 1/2 hr

total distance cover by A =1/2 * 30 miles = 15 miles

total distance by B = 1/2*40 miles = 20 miles

now from figure

$AB = \sqrt{(15^2+20^2)} = 25 miles$

$AB^2 =OA^2+ OB^2$

Differentiate above equation

$2AB\frac{dAB}{dt} = 2OA\frac{dOA}{dt} +2OB\frac{dOB}{dt}$

$25*\frac{dAB}{dt} = 15*30 +20*40$

$\frac{dAB}{dt} = \frac{1250}{25}$

$\frac{dAB}{dt} = = 50 miles/ hour$

Answer 2

Answer:

The rate of the distance between he and she is 50 miles/hr.

Explanation:

Given that,

Speed $\dfrac{dx}{dt}= 30\ miles/hr$

Speed $\dfrac{dy}{dt}= 40\ miles/hr$

After t hours,

He covered the distance x = 15 miles

She covered the distance y = 20 miles

We need to calculate the distance between he and she

Using Pythagorean theorem

$z^2=x^2+y^2$....(I)

Put the value into the formula

$z^2=15^2+20^2$

$z=\sqrt{15^2+20^2}$

$z=25\ miles$

We need to calculate the rate of the distance between he and she

From equation (I)

On differentiating of equation (I) w.r.to t

$z\dfrac{dz}{dt}=x\dfrac{dx}{dt}+y\dfrac{dy}{dt}$....(II)

Put the value in the equation

$\dfrac{dz}{dt}=\dfrac{15}{25}\times30+\dfrac{20}{25}\times40$

$\dfrac{dz}{dt}=50\ miles/hr$

Hence, The rate of the distance between he and she is 50 miles/hr.