Question

A person is standing 75 meters away from a kite and has a spool of string attached to the kite. The kite starts to rise straight up in the air at a rate of 2 m/sec and at the same time the person starts to move towards the kite's launch point at a rate of 0.75 m/sec. Is the length of the string increasing or decreasing after 4 secs and 20 secs?

Answer 1

you're asked about \[\frac{ d }{ dt } \sqrt{(75-x)^2 +y^2}\]

Answer 2

can you find that? in terms of x, y, dx/dt and dy/dt ?

(differentiate and treat x and y as functions of t: use the chain rule)

Answer 3

hello?

Answer 4

Yeah, do you derive the pythagorean theorem?

Answer 5

yes

Answer 6

I mean, probably. Not quite sure what you're asking tbh. Are you asking if you use the pythagorean theorem to find the length of the string in terms of x and y? If so the answer is yes.

Answer 7

Are you able to use the chain rule to differentiate that expression with respect to t?

Answer 8

Yeah.

Answer 9

Wouldn't it be dx/dt(2x)+dy/dt(2y)=d/dt(2d)? Then just plug in the values

Answer 10

yep you can do it that way...

Answer 11

as long as your 'x' is properly defined...

Answer 12

I already showed you the expression for 'd', so you can just differentiate that directly...

Answer 13

either way, same thing.

Answer 14

Wait, so x=75-x, right?

Answer 15

eh well... the horizontal side of the triangle is 75-x, that's correct. so in your expression that goes in where you put 'x' ... please don't say that x= 75-x though. It will just confuse you.

Answer 16

but yeah: d/dt (75-x)^2 + d/dt (y^2) =L* dL/dt

Answer 17

you have the right idea.

Answer 18

ok. got it so far?
the only other 'trick' is finding 'x' and 'y' at the given times... that's not really hard... you got that part?

Answer 19

Okay, so wouldn't dx/dt=0.75 and dy/dt=2? Since it's given in the equation or?

Answer 20

yep.

Answer 21

so \[\frac{ -(75-x)*(.75) + y*(2) }{ \sqrt{(75-x)^2 +y^2}} = \frac{ dL }{ dt }\]

Answer 22

So after you simplify, what would you do after?

Answer 23

no need to simplify... now it's just plugging in x and y at t=4 and then x and y at t=20

Answer 24

you got that part? or confused?

Answer 25

I don't get it.

Answer 26

x changes at a constant rate .75 m/s so after 4 seconds x is 3 m and 75 -x is 75-3 =72m
y also changes at a constant rate, so after 4 sec it's simply 2m/s *4 sec = 8m

Answer 27

same procedure for when t=20...

Answer 28

Oh okay, I see what you did now.

Answer 29

plug in the values and calculate dL/dt

Answer 30

Okay, I got it now thanks. :)

Answer 31

great:)