Solving for Distance: Ship A and B at 4pm

[Unknown9]

Homework Statement

At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm.

Homework Equations

??

The Attempt at a Solution

I was able to draw a picture of the problem, but I don't know where to go from there, and I also understand that I am trying to solve for dz/dt, but I have no clue how to do that, please help :).

 

[berkeman]

Homework Statement

At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm.

Homework Equations

??

The Attempt at a Solution

I was able to draw a picture of the problem, but I don't know where to go from there, and I also understand that I am trying to solve for dz/dt, but I have no clue how to do that, please help :).

Show us your work. Then we can offer tutorial hints. You need to do the work here at the PF.

 

[Unknown9]

Show us your work. Then we can offer tutorial hints. You need to do the work here at the PF.

What? You mean to tell me you won't do my work for me? :O

Here's what I got so far.
dx/dt = 35km/h,,, dy/dt = 25km/h,,, dz/dt = ?

Then we'll use the z² = x² + y²

Then we implicit differentiate it in regards to time.

2z(dz/dt) = dx(dx/dt) + 2y(dy/dt)

Solve for dz/dt

(dz/dt) = (1/z)(x*(dx/dt) + y*(dy/dt))

This is where I start to get a little confused... I know we plug in dy/dt and dx/dt into the equation, but what's z? and is dz/dt what I'm looking for?

 

[Unknown9]

So then I have (dz/dt) = (1/z) * (35x + 25y), so would the xyz be the distance? Like 150 as x and...?

 

[Unknown9]

Calculus1:Related rates

I'll give this a better title so people know what I'm talking about :).

Homework Statement

At noon, ship A is 150km west of ship B. Ship A is sailing east at 35km/h and ship B is sailing north at 25km/h. HOw fast is the distance between the ships changing at 4:00pm.

Homework Equations

x²+y² = z²

The Attempt at a Solution

Here's what I got so far.



dx/dt = 35km/h,,, dy/dt = 25km/h,,, dz/dt = ?

Then we'll use the z2 = x2 + y2

Then we implicit differentiate it in regards to time.

2z(dz/dt) = dx(dx/dt) + 2y(dy/dt)

Solve for dz/dt

(dz/dt) = (1/z)(x*(dx/dt) + y*(dy/dt))

This is where I start to get a little confused... I know we plug in dy/dt and dx/dt into the equation, but what's z? and is dz/dt what I'm looking for?

 

[Mark44]

You need to take into account the positions of both ships at time t. Let A_t and B_t be the coordinates of the two ships at time t. Let's suppose that ship A is at the origin at noon, so the coordinates of A₀ are (0, 0) and the coordinates of B₀ are (150, 0) at noon.

Ship A is moving east (to the right) at 35 km/hr, so what are its coordinates at an arbitrary time t? Ship B is moving north (upward) at 25 km/hr, so what are its coordinates at time t?

Get an equation for the distance between points A_t and B_t, then differentiate both sides of that equation, and evaluate it when t = 4 (4:00pm).

 

[blake knight]

You almost got it..bear in mind that noon means 12:00, so from noon to 4:00 the 2 ships traveled 4 hours already...continue your work from there.

 

[Mark44]

Unknown9,
You posted the same problem twice, which is not a good thing to do. The other thread you started is here: https://www.physicsforums.com/showthread.php?t=345299.

 

[berkeman]

Merged the two threads. Do not multiple post, Unknown9.

 

[Unknown9]

So I got z = sqr( (150-x)^2 + y^2)

Then I differentiated that and got.

dz/dt = 1/2( (150-x)^2 + y^2)^-1/2 * (-2(dx/dt)(150-x) + 2y(dy/dt))

when I plugged in the numbers, which I used x as 140 y as 100, and dx/dt as 35 and dy/dt as 25, I got

dz/dt = 1/(2*sqrt(10,100)) * 4300

am I close?

 

[blake knight]

I got the same result.