Question

A video camera is stationed 50 feet from a road that is part of a parade route to film the passing parade. Assume the parade floats are instructed to travel the route at a speed of 15 mi/hr.

a) At what rate must the camera rotate to follor the parade floats when they are 25 feet from the point P on the road closest ot the camera? at the instant they pass point p?

b) At what rate must the camera's focus adjust to film the floats clearly (that is, at what rate does the distance from the float to the camera change) when the floats are 200 feet from point P? when they are 50 feet from point P?

Answer 1

I'm just not sure how to set it up.

Answer 2

I'm awful with word problems.

Answer 3

Okay, so one leg of the triangle changes at 15 mi/hr. The other leg is constantly 50. The first question is asking the rate of change of the angle.

Answer 4

So write an equation relating the angle to the legs you know. The leg changing (we'll call it x) is opposite of the angle you care about, we'll call it angle a. Tan(a) = x/50

Answer 5

oh so maybe use trig like sintheta=opp/hypp or something like that?

Answer 6

is the road straight?

how far down the road do you start filming?

Answer 7

I don't know...

Answer 8

That doesn't matter. Don't worry.

Answer 9

But tan(a) = x/50, so solve for a. a = tan-1(x/50)

Answer 10

Take the derivative of a with respect to x.

Answer 11

(tan-1) is inverse tangent.

Answer 12

ok. I just don't understand the triangle in that problem
between the camera, road, and?

Answer 13

I'll draw a picture. Just a moment.

Answer 14

I am assuming that the line point p and the camera makes are perpendicular to the road, creating a right angle.

so it would be tan^-1 (25/50)

the other one would be tan^-1(200/50)-tan^-1(50/50)

Answer 15

ok so da/dt=1/sqrt(1+(x/50)^2)

Answer 16

times 1/50

Answer 17

oh ok. I get it.

Answer 18

Hope that picture clarifies.

Answer 19

times dx/dt...forgot that part.

Answer 20

Not quite right, I don't think. I have that derivative of arctan(x) is 1/(1+x^2). So derivative of arctan(x/50) would be (1/(1+x^2/50^2))*(1/50)

Answer 21

no dx/dt at the end?

Answer 22

The (1/50) is because of chain rule.

Answer 23

ok.

Answer 24

Yes, dx/dt also. That just gave you da/dx. You have to look at how quickly x is changing too.

Answer 25

ok.

Answer 26

ok

Answer 27

not sure how to do that?

Answer 28

I can't figure out a second equation that doesn't have too many variables.

Answer 29

Okay, so we have da/dx = 1/(50 + x^2/50)

Answer 30

So da/dt = da/dx * dx/dt, but dx/dt is given. It's 15 mi/hr.

Answer 31

i think it would be easier to convert that speed to ft per sec
15 mph = 22 ft/sec

Answer 32

oh ok, I meant x :P not dx/dt.

Answer 33

Good call, cow.

Answer 34

but I'm guessing for a its 25.

Answer 35

Yes. For a, it asks you to give da/dt when x = 25 and when x=0.

Answer 36

Do you see how to use the equation we derived to solve those two things?

Answer 37

i get
\[\frac{da}{dt} = \frac{(50)(22)}{x^{2} + 50^{2}}\]

Answer 38

also you need to know if they want it in radians per sec or degrees per sec

Answer 39

I got da/dt=(1/(50sqrt(1+(x/50)^2))(dx/dt)

Answer 40

I know, it looks super ugly.

Answer 41

isn't it already in radians?

Answer 42

Really not sure where your sqrt is coming from.

Answer 43

hmm yeah it should be, nevermind then

Answer 44

derivative of arctan(x) is 1/sqrt(1+x^2)

Answer 45

no sqrt, just 1/(1+x^2)

Answer 46

No it isn't?

Answer 47

woops?

Answer 48

Haha yes. Woops. You just used the wrong equation though. Everything else you did right.

Answer 49

lol ok. so its that but w/o the sqrt?

Answer 50

So your result should be da/dt = 1/(50 + x^2/50) * dx/dt

Answer 51

Yes. That turns out to be what you put except without the sqrt.

Answer 52

oh awesome!

Answer 53

thank you so much!

Answer 54

Need help on b?

Answer 55

(My pleasure) =)

Answer 56

yes please.

Answer 57

Okay, so back to the picture, call that blue line h, for hypotenuse. We want to know how quickly it is changing with respect to time for the given x values.

Answer 58

Make sense?

Answer 59

yes.

Answer 60

so opposite is 200 ft.

Answer 61

So write an equation relating h to x.

Answer 62

kk

Answer 63

sintheta=200/h

Answer 64

so theta=(sin-1)(200/h)

Answer 65

Except that theta is going to be changing along with x, which would give us 2 variables. Can we write it more simply? Write an equality between h and x so that x is the only variable?

Answer 66

If I give you x, how could you find h?

Answer 67

oh duh h^2=50^2+200^2

Answer 68

thats correct but you dont want to use 200 yet, that is an x value you will plug in at the end
you are looking for dh/dt
you solved for da/dt
dh/dt = da/dt * dh/da
get h in terms of a first

Answer 69

oh ok.

Answer 70

Yes, except keep it as x for now.

Answer 71

ok.

Answer 72

So h = sqrt(50^2 + x^2). Now find dh/dx.

Answer 73

ok

Answer 74

give me a sec

Answer 75

ok. dh/dt=(2x/2h)(dx/dt)

Answer 76

oh ok im wrong then, you only need dh/dx not dh/dt

Answer 77

oh ok.

Answer 78

Um... no you do need dh/dt. But first you should find dh/dx.

Answer 79

so dh/dx=2x/2h

Answer 80

But I'm not sure how you got that...

Answer 81

h^2=50^2+x^2

Answer 82

Derivative with respect to x of sqrt(50^2 +x^2)

Answer 83

2h(dh/dx)=0+2x

Answer 84

dh/dx=2x/2h

Answer 85

First solve for h, so sqrt both sides.

Answer 86

did I do something wrong? I was trying to not use the chain rule lol.

Answer 87

Yeah, that's not quite right. It doesn't work that way.

Answer 88

Okay, so you got to h^2 = 50^2 + x^2 and then tried to derive, but d/dx of h^2 is not 2h.

Answer 89

d/dh of h^2 is 2h, but not d/dx.

Answer 90

k. so dh/dx=1/2(2500+x^2)^-1/2(2x)

Answer 91

See?

Answer 92

oh ok.

Answer 93

Yes. And (2500 + x^2)^-1/2 is 1/sqrt(2500 + x^2)

Answer 94

ok.

Answer 95

So simplifying gives you 2x/(2*sqrt(2500+x^2))

Answer 96

can I plug in x now?