calculus problem

**samsum** · November 2nd 2006, 09:11 AM

a police cruiser, approaching a right-angled intersection from the north, is chaing a speeding car that has turned the corner and is now moving straight east. When the cruiser is 0.6 miles north of the intersection and the car is 0.8 miles to the east of the intersection. The police determine with radar that he direct distance between them and the car is increasing at 20 mph. If the cruiser is moving at 60 mph at the instant of measurement. what is the speed of the car?

**Soroban** · November 2nd 2006, 10:13 AM

Hello, samsum!

An interesting problem . . .

A police cruiser, approaching a right-angled intersection $A$ from the north,
is chasing a speeding car that has turned the corner and is now moving straight east.
When the cruiser is 0.6 miles north of $A$ and the car is 0.8 miles east of $A$ ,
the police determine that the direct distance between them and the car is increasing at 20 mph.
If the cruiser is moving at 60 mph at that instant, what is the speed of the car?

Code:

      |
    P *
      |  *
      |     *   z
    y |        *
      |           *
      |              *
      * - - - - - - - - *
    A          x        C

The police car is at $P,\:y$ miles north of $A.$

The car is at $C,\:x$ miles east of $A.$

The distance between them is: . $z^2 \:=\:x^2 + y^2$

Differentiate with respect to time: . $2z\!\cdot\!\frac{dz}{dt} \;= \; 2x\!\cdot\!\frac{dx}{dt} + 2y\!\cdot\frac{dy}{dt}$

. . Divide by 2: . $z\!\cdot\frac{dz}{dt}\;=\;x\!\cdot\frac{dx}{dt} + y\!\cdot\!\frac{dy}{dt}$ [1]

We are told that: . $x = 0.8,\;y = 0.6,\;\frac{dz}{dt} = 20,\;\frac{dy}{dt} = -60$
. . (The distance $y$ is decreasing.)

When $x = 0.8$ and $y = 0.6$ , the hypotenuse $z = 1.$

Substitute into equation [1]: . $1(20) \;= \;(0.8)\!\cdot\!\frac{dx}{dt} + (0.6)(-60)$

And we have: . $(0.8)\!\cdot\frac{dx}{dt} \:=\:56\quad\Rightarrow\quad \frac{dx}{dt} = 70$

Therefore, the speed of the car is $70\text{ mph.}$

. . . BUS-ted!

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