Question

A road running north to south crosses a road going east to west at the point P. Car A is driving north along the first road, and car B is driving east along the second road. At a particular time car A is 10 kilometers to the north of P and traveling at 80 km/hr, while car B is 15 kilometers to the east of P and traveling at 100 km/hr. How fast is the distance between the two cars changing?

Answer 1

This is just like the last problem.

Answer 2

Almost exactly.

Answer 3

really?

Answer 4

Yes. We can follow the same method.

Answer 5

I really hate word problems.

Answer 6

They trip me up

Answer 7

ok so y=10+100t, x=15+100t

Answer 8

and dx/dt=100km/hr?

Answer 9

Almost. Check your speeds.

Answer 10

oh I didn't see the 80

Answer 11

y=10+80t

Answer 12

For reference.

Answer 13

Yes. Correct. And what is the formula for the distance d?

Answer 14

d=sqrt(x^2+y^2)

Answer 15

Mhmm. So what's next?

Answer 16

take the derivative

Answer 17

of the distace according x?

Answer 18

Of the distance according to t

Answer 19

t?

Answer 20

Not yet. How about substituting to get an equation for d in terms of t?

Answer 21

oh yes!

Answer 22

We have d = sqrt(x^2 + y^2)
x = 15 + 100t
y = 10 +80t

Answer 23

yes. simplifying now.

Answer 24

Good good =) I'm not gonna check your work on this one.

Answer 25

k so d=sqrt(325+4600t+16400t^2)

Answer 26

then take dd/dt

Answer 27

Yup. =)

Answer 28

You can not substitute also..

\[\frac{dD}{dt} = \frac{d}{dt}[\sqrt{x^2 + y^2}]\] \[=\frac{1}{2\sqrt{x^2 + y^2}}\cdot \frac{d}{dt}(x^2 + y^2)\]\[=\frac{1}{2\sqrt{x^2 + y^2}}\cdot(2x\frac{dx}{dt} + 2y\frac{dy}{dt})\]

Makes for a little less work actually.

Answer 29

I'm confused now.

Answer 30

Ah. Good point, polpak. Might be simpler for her to understand the other way though.

Answer 31

Sorry, feel free to ignore me if you like.

Answer 32

Yeah, what you're doing is right, Nasryn. He just has another method.

Answer 33

Did you get the derivative? It's still going to have t involved like last time.

Answer 34

Should get the same answer either way. Your way just required you to integrate then differentiate again. My way is just differentiate then plug/chug.

Answer 35

yes. dd/dt=(16400t+2300)/sqrt(325+4600t+16400t^2)

Answer 36

look ridonkulous

Answer 37

Very good. And what is t?

Answer 38

0, right?

Answer 39

I'm not sure.

Answer 40

I have to go, sorry. I'm going with friends to see some live jazz music =)

Answer 41

ok no worries.

Answer 42

But t is 0, so if you plug in 0, that should give you your answer =)

Answer 43

:) thanks.

Answer 44

Later.

Answer 45

\[\frac{dD}{dt} = \frac{1}{2\sqrt{x^2 + y^2}}\cdot(2x\frac{dx}{dt} + 2y\frac{dy}{dt})\]
\[\implies \frac{dD}{dt}|_{x=15, y=10} = \frac{1}{2\sqrt{15^2 + 10^2}}\cdot(2(15)(100) + 2(10)(80))\]\[= \frac{1}{2\sqrt{325}}\cdot 2(1500 + 800)\]\[=\frac{2300}{5\sqrt{13}}\]\[={460 \over \sqrt{13}}\]

Answer 46

:D wonderful thank you.

Answer 47

You can at least check your answer if you don't follow my method, and/or you can try to figure out how I did it ;)

Basically I'm just implicitly differentiating and using the chain rule here.

Answer 48

yeah, I did it your way because I'm pretty sure that's the way I'm supposed to do it. So thank you!