1. ## Comparing Related Rates

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A team has been working to convert diesel-powered cars to run just as efficiently on used cooking oil! They want to compare the mileage and speed of their prototype with that of the diesel-powered car.

The prototype is 100 meters south of an intersection, while the diesel car is 100 meters east of the intersection. Both vehicles start moving at the same time. The prototype moves north, toward the intersection, and the diesel car moves east, away from the intersection. If the prototype is traveling at a velocity of 3 meters per second and the diesel car is traveling at 2 meters per second, what is the rate of change of the distance between the cars after four seconds? Round off your answer to two decimal places.
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I've been trying to do this for a very long time, yet the answer is still wrong, here's pretty much what I tried:

Prototype's Distance, P = 100 m.
Prototype's Speed, dP/dt = - 3 m./sec. (Negative because the distance from the intersection is decreasing)
Diesel's Distance , D = 100 m.
Diesel's Speed, dD/dt = 2 m./sec.
Time, t = 4 secs.
Distance Between Cars = d

Find dd/dt:

Distance D travels in 4 secs. = D2, so

D2 = D + (dD/dt)(4)
D2 = 100 + [2(4)]
D2 = 100 + 8
D2 = 108

Distance P Travels in 4 secs. = P2, so

P2 = P + (dP/dt)(t)

Since distance is always positive, dP/dt is positive, so

P2 = 100 + [3(4)]
P2 = 100 + 12
P2 = 112

d² = (D2)² + (P2)²
d² = (108)² + (112)²
d² = 11664 + 12544
d² = 24208
d = √24208
d = 155.58

Differentiating Implicitly Over Time,

2d(dd/dt) = 2(D2)(dD/dt) + 2(P2)(dP/dt)
d(dd/dt) = (D2)(dD/dt) + (P2)(dP/dt)
155.58(dd/dt) = 108(2) + 112(- 3)
155.58(dd/dt) = 216 - 336
155.58(dd/dt) = - 120
dd/dt = - 120 / 155.58
dd/dt = - 0.7713

Any ideas? am I doing something wrong?

2. Originally Posted by icanthearyou
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A team has been working to convert diesel-powered cars to run just as efficiently on used cooking oil! They want to compare the mileage and speed of their prototype with that of the diesel-powered car.

The prototype is 100 meters south of an intersection, while the diesel car is 100 meters east of the intersection. Both vehicles start moving at the same time. The prototype moves north, toward the intersection, and the diesel car moves east, away from the intersection. If the prototype is traveling at a velocity of 3 meters per second and the diesel car is traveling at 2 meters per second, what is the rate of change of the distance between the cars after four seconds? Round off your answer to two decimal places.
-----------------------

I've been trying to do this for a very long time, yet the answer is still wrong, here's pretty much what I tried:

Prototype's Distance, P = 100 m.
Prototype's Speed, dP/dt = - 3 m./sec. (Negative because the distance from the intersection is decreasing)
Diesel's Distance , D = 100 m.
Diesel's Speed, dD/dt = 2 m./sec.
Time, t = 4 secs.
Distance Between Cars = d

Find dd/dt:

Distance D travels in 4 secs. = D2, so

D2 = D + (dD/dt)(4)
D2 = 100 + [2(4)]
D2 = 100 + 8
D2 = 108

Distance P Travels in 4 secs. = P2, so

P2 = P + (dP/dt)(t)

Since distance is always positive, dP/dt is positive, so

P2 = 100 + [3(4)]
P2 = 100 + 12
P2 = 112

d² = (D2)² + (P2)²
d² = (108)² + (112)²
d² = 11664 + 12544
d² = 24208
d = √24208
d = 155.58

Differentiating Implicitly Over Time,

2d(dd/dt) = 2(D2)(dD/dt) + 2(P2)(dP/dt)
d(dd/dt) = (D2)(dD/dt) + (P2)(dP/dt)
155.58(dd/dt) = 108(2) + 112(- 3)
155.58(dd/dt) = 216 - 336
155.58(dd/dt) = - 120
dd/dt = - 120 / 155.58
dd/dt = - 0.7713

Any ideas? am I doing something wrong?
Hint:

s = $\sqrt{p^2 + q^2}$

$\dfrac{ds}{dt} = \dfrac{p\dfrac{dp}{dt}+q\dfrac{dq}{dt}}{\sqrt{p^2 + q^2}}$

3. Thanks for the response. :]
But now I'm more confused than before lol

If you don't mind, can you explain a bit please?

4. Hello, icanthearyou!

I assume you made a sketch.

The prototype is 100 meters south of an intersection $O$,
. . while the diesel car is 100 meters east of $O.$
Both vehicles start moving at the same time.
The prototype moves north, toward the intersection;
. . the diesel car moves east, away from the intersection.
The prototype is traveling at 3 m/sec and the diesel car is traveling at 2 m/sec.
What is the rate of change of the distance between the cars after four seconds?

Code:
     O     100      B   2t    D
-   * - - - - - - * - - - - *
:   |                 *
:   |100-3t     *
100  |     *   x
: P *
:   |
:   |3t
:   |
-   *
A
The prototype starts at $A:\;OA = 100$
In $t$ seconds, it has moved $3t$ feet to point $P:\;OP = 100-3t$

The diesel starts at $B:\;OB = 100$
In $t$ seconds, it has moved $2t$ feet to point $D:\;OD = 100 + 2t$

In right triangle $POD$, we have: .$PD^2 \:=\:OP^2 + OD^2$
Let $x$ equal the distance $PD.$

Then we have: .$x \;=\;\sqrt{(100-3t)^2 + (100 + 2t)^2}$
. . which simplifies to: .$x \;=\;(13t^2 - 200t + 20,\!000)^{\frac{1}{2}}$

Differentiate with respect to time:

. ..$\dfrac{dx}{dt} \;=\;\frac{1}{2}(13t^2 - 200t + 20,\!000)^{-\frac{1}{2}}(26t - 200)$

. . $\dfrac{dx}{dt} \;=\;\dfrac{13t-100}{\sqrt{13t^2 - 200t + 20,\!000}}$

Now let $t = 4.$

5. Hello Soroban,

Great and detailed post and I think I pretty much understand the concept.
Thank you both for the help :]

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