Car

[legacyofpiracy]

Hello again, I was wondering if I could have some help on this related rates problem

Car A is traveling towards a gas station at a speed of 25mph. Car B is moving away from the station at 20mph. Let X be the distance between Car A and the Station at time t, let y be the distance between Car B and the Station at time t.

1. find the distance and rate of change between the two cars when x=4, y=3
(i keep on wanting to just do the pythagorean theorem to find the distance...but it seems to straightforward...am I right?)
2. (There is actually a theta sign at the car A angle) and I need to find the rate of change of the theta in radians per hour again when x=4 and y=3

 

[galactus]

Without seeing your diagram, I will take a stab at it.

For #1:

Let D be the distance between the cars. We want \(\displaystyle \frac{dD}{dt}\) when

x=4 and y=3. We know \(\displaystyle \frac{dy}{dt}=20\) and \(\displaystyle \frac{dx}{dt}=-25\)

\(\displaystyle D^{2}=x^{2}+y^{2}\) [1]

\(\displaystyle D^{2}=3^{2}+4^{2}\) implies that D=5.

Differentiate [1] with respect to t and plug in the values you know, solve for \(\displaystyle \frac{dD}{dt}\)

 

[legacyofpiracy]

arg sorry about the picture...i'm having trouble with it

basically it is just a standard right triangle with the hypotenuse on the right. Car B is the top (left) angle, the station is directly below it (the bottom left 90degree angle) and Car A is to the right of that (bottom right angle)

Car B however is moving away from the station as Car A is approaching it, so obviously the triangle lengthens in height and looses its width over time

 

[legacyofpiracy]

Oh wonderful thank you, so I was right about the pythagorean theorem at least . Thanks that makes perfect sense, I was actually headed in the right direction but I just wanted to clarify. What I am, however, puzzled on is the 2nd question - how to find the rate of change of theta (in car A's angle)

 

[legacyofpiracy]

Ok I think I might be vaugely around what I am supposed to be doing for part 2, does this look right anyone?

Ok so i need to find the rate of change of theta when x=4 and y=3

so i decided to use sin(theta)=opposite/hypotenuse
and since the Y value is given as 3 and I figgured out the distance (hypotenuse) to be 5 I could set up this equation and take the derivative and solve for Dtheta/dt

sin(theta)=3/5
5sin(theta)=3

derivative

5cos(theta)*(Dtheta/dt)=0
but then I figgure this can't be right because solving for Dtheta/dt would just give me 0....

any help please?

 

[Gene]

y = 3+20t
x = 4-25t
tan(\(\displaystyle \Theta\)) = y/x
\(\displaystyle \Theta\) = arctan(y/x) =
arctan((3+20t)/(4-25t))
d\(\displaystyle \Theta\) = d(arctan((3+20t)/(4-25t)))

d(arctan(u) = du/(1-u²)
u = (3+20t)/(4-25t)

 

[legacyofpiracy]

sorry but I am a bit confused by your variables. First of all what is "t" in the 3+20t and why are you subtracting it from the Y=3? Also when you get down to taking the derivative (at least that is what i am assuming the d to be) of the arctan(u) what is the "u" ? Sorry about this, I am just having trouble understanding your reasoning here :?

 

[Gene]

I was told

at time t

so I used t for time.

u is a general variable used in most calculus texts. As in
du²=2u*du
so when you want
d(x+5)² you substitute x+5 for u and
get 2(x+5)*d(x+5).
u is defined in the last line.

y = 3+20t and
x = 4-25t
are the equations for where cars B and A are at time t. B left the station (y=0) 9 minutes ago if any one is interested.