How Fast is the Distance Changing Between Two Ships and a Man's Shadow?

[aMo0oDi]

Hi guys.. i was studying calculus and i found some problems in solving these two questions.. would you please help me to figure out the answers?


1) At noon ship A is 150 km west of ship B. Ship A is sailing East at 35 km/h and ship B is sailing North at 25 km/h. How fast is the dstance between the ships changing at 4 pm?

2) a spotlight on the ground shines on a wall 12 meters away. if a man 2 meters tall walks for the spotlight toward the building at a speed of 1.6 meter per second, how fast is the length of his shadow on the building decreasing when he is 4 meters from the building?

 

[AKG]

1) Think of ship A traveling right on the x-axis, starting at the point (-150,0). Think of ship B traveling up on the y-axis, initially at (0,0). Let A represent ship A's distance from the origin, and B represent B's distance from the origin. The distance between the two ships is:

\sqrt{A^2 + B^2}

If you think about it, place ship A at any point on the x-axis and B at any point on the y-axis, and you'll see the distance between the ships is the hypoteneuse of a triangle, whose arms are the segment from the origin to ship A, and the segment from the origin to ship B. and given the lengths of two arms of a right triangle, you know how to find the length of the hypoteneuse by Pythagorean Theorem, which is how I got the equation above. Now, how do we express A and B in terms of time? Let t represent time, in hours.

A = -150 + 35t
B = 0 + 25t

You should be able to see clearly why the equations above are right. If not, maybe try plotting the position of each ship after a bit of time goes on, and notice how the distances to the origin are properly described by those equations. Now, going back to the distance, let's call it D:

D = \sqrt{A^2 + B^2}

D = \sqrt{(35t - 150)^2 + (25t)^2}

D = \sqrt{1850t^2 - 10500t + 22500}

Now, to find the rate of change of distance with respect to time, differentiate D with respect to t:

dD/dt = \frac{3700t - 10500}{2\sqrt{1850t^2 - 10500t + 22500}}

To find the rate of change of distance at 4 p.m., that is 4 hours later, plug in t=4, and you'll have your answer.

 

[Phymath]

well its been awhile sense I've done these but...draw the orginal situation and u find ship A's equation of motion is x(t) = 35t - 150 and B's is y(t) = 25t

the distance between two points in a flat plane is d = sqrt{(x - x_o)^2 -(y - y_0)^2}, but u see A's motion is only on the x axis... so x_0 = 0 and... also for B y_0 = 0 so sub the eq's for x and y u get...

d = sqrt{(35t - 150)^2 + (25t)^2}

dd/dt = change of distance/change of time

dd/dt = 1250t + 70(35t-150)/(2 sqrt(625t^2 + (35t - 150)^2)) and t = 4 so the answer...4,996.517 km/h

i have no idea if this is right lol but check it out lol

 

[AKG]

I got an answer of something like 21.4 km/h. Did you really get something like 5000 km/h? How could they possibly be moving away from each other at that rate? Even if they were headed in opposite directions they could only be separating at a rate of 25+35=60 km/h, right?

 

[rumaithya]

AKG, I think your answer sounds correct.

 

[aMo0oDi]

thanks a lot for helping us figuring out the way of solving the first problem.. what about the second one?

 

[aMo0oDi]

2) a spotlight on the ground shines on a wall 12 meters away. if a man 2 meters tall walks from the spotlight toward the building at a speed of 1.6 meter per second, how fast is the length of his shadow on the building decreasing when he is 4 meters from the building?

 

[aMo0oDi]

i got 1.01 m/s is that correct ..
I did it by using x^2+y^2= z^2
x=8 y= 2
i found the hyp as equal to 8.24 m b
soo i used it to get the time which is 7.5 s T=d/s T=12/1.6
then i used the time with the hyp to get the speed . which is s=d/t
8.24/7.5 = s
im wondering is that correct ??

 

[keltix]

http://calclab.math.tamu.edu/~belmonte/m151/L/c3/L3A.pdf