find the distance in kilometers between ship A and ship B?

Question

Ship A is traveling west toward a light house at 15 km/hr. Ship B is traveling north away from lighthouse at 10km/hr. x is distnace between ship A and lighthouse at time t, and y is distance between ship B and lighthouse at time t.

a. find the distance in km, between ship A and Ship B if x=4 and y=3. ((do i use pythagorean thm?)
b. Find rate of change in km/hr of the distance between the two ships if x = 4km and y=3 km. (Does the distance between the two ships get smaller??)
c. let theta be the angle wit h a vertex of the angle formed between light house and ship B (ship A as vertex). find the rate of change in theta, in radians per hour when x = 4 km and y = 3km.


THANKS so much for any help at all, you dnt nee to do them all to help, just any help is GREATLY appreciated! :)

what equation?
i dont have an equation???
i used the pythagorean thm and took the derivative of it, and just plugged in all the values i had (all except for the rate of change).
is that correct?

well for the last part, its confusing, but think of the drawing of the triangel with the ships--they form a right triangle, and ship a is east of it light house, and ship b is north of light house. the angle it is talking about is that of ship A (As ship a gets closer to light house, what is the rate in which it is chaning when x =4 and y=3. )
i used the tan(theta)=x/y, and took the derivative of it? but then i got confused.
hope this helps for the last part.

Answer 1

a) yes, u do use the pythagorean theorem, since the 2 ships are going in perpendicular directions. this gives you:
d=sqrt(3^2+4^2)=sqrt(9+16)=5
b) this is a typical related rates problem. first, u must write equations for x and y. here, y=10t, where t is the elapsed time in hours, b/c the distance between B and the lighthouse is increasing at 10 km/hr. also, x=x0-15t, where x0 is the initial distance between A and the lighthouse, b/c the distance is starting at some initial value and decreasing at 15 km/hr. since we know what y is, we can use it to solve for t:
t=y/10
=3/10 hr
we know what x is, so we can plug in the values for x and t to find x0:
4=x0-15*3/10
=x0-4.5
x0=9.5 km
now, x=9.5-15t and y=10t. the distance between the ships can be found using the pythagorean theorem, so:
s=sqrt(x^2+y^2)
=sqrt((9.5-15t)^2+(10t)^2)
=sqrt(90.25 -285t +225t^2 +100t^2)
=(90.25-285t+325t^2)^.5
to find the rate of change of the distance, take the derivative of s with respect to t:
ds/dt=d((90.25 -285t +325t^2)^.5)/dt
=.5*(90.25 -285t +325t^2)^-.5* d(90.25 -285t +325t^2)/dt
=(-285 +650t)/ (2*sqrt(90.25- 285t +325t^2))
now, plug in t=3/10 hr:
ds/dt=(-285 +650(3/10))/ (2*sqrt(90.25 -285(3/10) +325(9/100)))
=(-285 +195)/ (2*sqrt(90.25 -2850 +32500))
=-90/(2*sqrt(34))
=-7.7 km/hr
(u might want 2 check the work in b) & c) yourself 2 see if it's correct).
c) now, u have 2 do some visualization. the tangent of the angle you're given is equal to y/x, so:
tan(theta)=y/x
=10t/(9.5 -15t)
now, u must take the derivative of both sides with respect to t:
d(tan(theta))/dt=d(10t /(9.5 -15t))/dt
sec(theta)^2*dtheta/dt= (d(10t)/dt *(9.5 -15t)- d(9.5 -15t)/dt *10t)/ (9.5- 15t)^2
=10(9.5-15t) +15(10t) /(9.5-15t)^2
=95 -150t +150t /(9.5-15t)^2
=95 /(9.5-15t)^2
you know what y and x equal, so use the formula to find theta:
theta=arctan(y/x)
=arctan(3/4)
=.64 radians
you also know that t=3/10, so plug t and theta back into the equation to solve for dtheta/dt:
sec(36.9)^2 *dtheta/dt=95 /(9.5 -15(3/10))^2
1.5625* dtheta/dt=95 /(9.5-4.5)^2
=95 /(4^2)
=95 /16
=5.9375
dtheta/dt=3.8

Answer 2

a) Yes use the Theorem of Pythagarus here.
d^2 = x^2 + y^2
d^2 = 3^2 +4^2 = 25

b) --a little trickier, but since you need only a discreet value at a discreet time, the relative velocity will be given by:
v^2 = 15^2 + 10^2 @ α = arctan(10/15) with respect to A, (0° = East) (From A, B appears to be going East)
(Since 10/15 < 4/3, the ships are closing at this moment)
(vr = 10*4/5 - 15*3/5) = -1 kph

c) Again, looking for an instantaneous value:
With respect to A,
the tangential component of B's velocity
vt = -10*3/5 - 15*4/5 (+ being counterclockwise)
vt = -(30+60)/5 = -90/5 = -18 kph
vt = rω
ω = vt/r
ω = -18/5 = -3.6 rad/hr

Answer 3

a) yes d = sqrt(4^2 + 3^2)

b) Yes, at this point the distance is getting smaller.
Since d^2 = x^2 + y^2, (d/dt) d^2 = (d/dt)x^2 + (d/dt)y^2 = 2x(dx/dt) + 2y(dy/dt) = 2*4*15 + 2*3*-10 = 60

So, (d/dt)d^2 = 60 and (d/dt)d = sqrt 60 = +- 7.746 km/hr (use - sign)

c) Gotta think about this some more!

Answer 4

A) Pythagorean
B) No
C) You have the formula, just plug in the numbers.

Your equation does not seem to account for the curvature of the earth........

Answer 5

for first question yes u r right u use pytha therom

second question is about diff between velocitis

third i couldent understand

bye