Rate of change problem please hellp

[HousEDM]

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[stapel]

For other users, the text at the link states as follows:

A boat leaves a dock at 2:00 PM and travels due south at a speed of 20 km/hr. Another boat has been heading due east at 15 km/hr and reaches the same dock at 3:00 PM. How many minutes past 2:00 PM were the boats closest together?

To the poster:

I have no idea what your "picture" is meant to mean. So let's try using what you learned back in algebra. Draw a dot for the dock. Draw a line going downward from this dot; this represents the path of the first boat. Draw a dot on this line, below the dot for the dock; this dot represents the position of the first boat. Draw another line horizontally through the dot; this represents the path of the second boat. At 2:00 PM, the second boat was still west of the dock, so draw a dot on the horizontal line, with this dot being to the left of the dot for the dock and representing the position of the first boat. Draw a line joining the dots for the two boats; this represents the distance between the boats. This line forms a right triangle, the third vertex being the dot for the dock.

Given that the first boat is drawing away from the dock at 20 km/hr (or 1/3 km per minute), what expression, in terms of minutes of time "t", stands for the distance of the first boat from the dock? Given that the second boat is approaching the dock at 15 km/hr (or 1/4 km per minute), what expression stands for the distance of the second boat from the dock?

Apply the Pythagorean Theorem to the right triangle in order to obtain your equation for the distance, D, of the two boats in terms of time t.

Now turn to calculus: You are given rates of speed, which are of course derivatives of position. These rates are with respect to time, so the derivatives will be with respect to "t". What do you get when you differentiate?

Please be complete. Thank you! :wink:

 

[HousEDM]

For other users, the text at the link states as follows:


To the poster:

I have no idea what your "picture" is meant to mean. So let's try using what you learned back in algebra. Draw a dot for the dock. Draw a line going downward from this dot; this represents the path of the first boat. Draw a dot on this line, below the dot for the dock; this dot represents the position of the first boat. Draw another line horizontally through the dot; this represents the path of the second boat. At 2:00 PM, the second boat was still west of the dock, so draw a dot on the horizontal line, with this dot being to the left of the dot for the dock and representing the position of the first boat. Draw a line joining the dots for the two boats; this represents the distance between the boats. This line forms a right triangle, the third vertex being the dot for the dock.

Given that the first boat is drawing away from the dock at 20 km/hr (or 1/3 km per minute), what expression, in terms of minutes of time "t", stands for the distance of the first boat from the dock? Given that the second boat is approaching the dock at 15 km/hr (or 1/4 km per minute), what expression stands for the distance of the second boat from the dock?

Apply the Pythagorean Theorem to the right triangle in order to obtain your equation for the distance, D, of the two boats in terms of time t.

Now turn to calculus: You are given rates of speed, which are of course derivatives of position. These rates are with respect to time, so the derivatives will be with respect to "t". What do you get when you differentiate?

Please be complete. Thank you! :wink:

But how can you know the equation for distance D, we don't know at which point the boats are closes therefore it would only b D^2=a^2+b^2

 

[HallsofIvy]

But how can you know the equation for distance D, we don't know at which point the boats are closes therefore it would only b D^2=a^2+b^2

Yes, that's the whole point! Differentiating both sides with respect to time, t, we have \(\displaystyle 2D\frac{dD}{dt}= 2a\frac{da}{dt}+ 2b\frac{db}{dt}\) or, dividing both sides by 2, \(\displaystyle \frac{dD}{dt}= a\frac{da}{dt}+ b\frac{db}{dt}\). The boats are "closest together" when \(\displaystyle \frac{dD}{dt}= 0\).

 

[HousEDM]

solving this gives 0=-20a +15b , we don'tknow the values of either of the two

 

[Ishuda]

Another approach: we can take the origin as the dock (0,0) and do so.
Boat 1 at 20 km/h is at (x1(t), y1(t)) at time t. Since it was at the dock at time 0 (2:00 PM) and is traveling due south at 20 km/hr,
x1(t) = 0 <<<<<<highlight for answer
y1(t) = -20 t <<<<<<highlight for answer
where t is in hrs and x1 and y1 are in km.

Boat 2 at 15 km is at (x2(t), y2(t)) at time t. Since it was 15 km due west of the dock at t=0 [=2:00 pm = one hour before 3:00 PM] and is traveling 15 km/hr due west
x2(t)=15 (t-1) <<<<<<highlight for answer
y2(t)= 0 <<<<<<highlight for answer
where t is in hrs and x2 and y2 are in km.

At time t they are D km apart where
D = \(\displaystyle \sqrt{[x1(t) - x2(t)]^2 + [y1(t) - y2(t)]^2}\)

 

[HousEDM]

Another approach: we can take the origin as the dock (0,0) and do so.
Boat 1 at 20 km/h is at (x1(t), y1(t)) at time t. Since it was at the dock at time 0 (2:00 PM) and is traveling due south at 20 km/hr,
x1(t) = 0 <<<<<<highlight for answer
y1(t) = -20 t <<<<<<highlight for answer
where t is in hrs and x1 and y1 are in km.

Boat 2 at 15 km is at (x2(t), y2(t)) at time t. Since it was 15 km due west of the dock at t=0 [=2:00 pm = one hour before 3:00 PM] and is traveling 15 km/hr due west
x2(t)=15 (t-1) <<<<<<highlight for answer
y2(t)= 0 <<<<<<highlight for answer
where t is in hrs and x2 and y2 are in km.

At time t they are D km apart where
D = \(\displaystyle \sqrt{[x1(t) - x2(t)]^2 + [y1(t) - y2(t)]^2}\)

From there finding the derivative in terms of t would be near impossible no?

 

[HallsofIvy]

Start over- because we take a to be the distance the first boat is east of the initial position, and the boat is going east at constant speed 15 km/hr, then a= 15t. Since we take b to b the distance the second boat is south of the initial position, and the boat is going south at constant speed 20 km/hr, b= 20t. So \(\displaystyle D^2= a^2+ b^2= 225t^2+ 400t^2= 625t^2\). Differentiate that with respect to t.

 

[Ishuda]

From there finding the derivative in terms of t would be near impossible no?

No, actually it is rather a simple function. Actually pretty close to what HallsOfIvy has.

 

[Ishuda]

Start over- because we take a to be the distance the first boat is east of the initial position, and the boat is going east at constant speed 15 km/hr, then a= 15t. Since we take b to b the distance the second boat is south of the initial position, and the boat is going south at constant speed 20 km/hr, b= 20t. So \(\displaystyle D^2= a^2+ b^2= 225t^2+ 400t^2= 625t^2\). Differentiate that with respect to t.

But that is relative to its own time, i.e. a is 15 km east of where it was at time 2:00 PM but it wasn't at the dock at that time so the distance is offset back to 2:00PM for a point at time equal zero. You could also approach it as b is 20 km south of where it was at 2:00 PM but, if you want to use 3:00 PM as you zero time, boat 1 must be offset the hour from 2:00 when it was at the dock.

I'm contused so, anyway, make a question of that and answer it.

 

[stapel]

solving this gives 0=-20a +15b , we don'tknow the values of either of the two

You seem to be waiting for somebody to provide you with the completed solution. Only by doing will you learn how to do. Did you follow any of the step-by-step instructions provided in the first reply?

Using algebra, the rate 1/3 kpm, and the variable t for time, what expression stands for the distance covered in t minutes at 1/3 kpm? This is the expressions for the length of one of the legs.

Using algebra, the rate 1/4 kpm, and the variable t for time, what expression stands for the distance covered in t minutes at 1/4 kpm? This is the expressions for the length of the other of the legs.

Using these expressions and the Pythagorean Theorem, what expression stands for the distance between the two boats after t minutes?

Please reply showing your work and results for these questions. Thank you!