Question

ok, last question in the assignment that i can't seem to get, [derivative = hard ;_;], once again the question is attached in next post.

Answer 1

alright, what i know so far:
after 3 seconds, the x[t] = 42ft and the s[t]=58.7 while the y[t] is given as 41ft.
so now im not exactly sure what the question wants cause wouldnt the rate of s[t] remain the same after 3 seconds? o:

Answer 2

this is the same as the problem you posted earlier, Instead of the diagonal of the rectangle, you have the hypotenuse of the triangle in this problem

Answer 3

ya, i know that, but the question is asking "how fast is the distance s[t] between bike and balloon increasing 3 seconds later".... so i mean, won't the s[t] have the same rate increasing after 3 secs? why would the rate change?

Answer 4

alright, i see a mistake i made... the 41ft i used to fine the s[t] i didnt account that after 3 secs it would also increase like the x[t] is... so the y[t] after 3 secs would be 56ft, the x[t] 42ft and the s[t] = 70

Answer 5

yes.

Answer 6

alright, so now i calculated the s[t], x[t] and y[t] would be after 6 secs and i got, y[t] = 86ft, the x[t] = 126ft and s[t] = 153. So using these, i did: 153-70/6-3 = 27.5ft/sec and i believe this is the rate s[t] is increasing by.

Answer 7

soooo, is this correct? o:

Answer 8

how did you get x(t) = 126 ft for t =6sec?

Answer 9

um, at 3 sec the x[t] is = 42ft so at 6 sec its 14ft/sec times 6s = 84 ft + 42 ft = 126 ft

Answer 10

Hold on. Do you mean to say that If i am travelling at a constant rate of 1ft/sec, I will cover 3 feet in 3 seconds and 3+6 = 9 feet in 6 seconds?

Answer 11

OHHHHHHHHHHHHHHHHHH, ok ok ok... my bad. i see now, k.
so the x[t] after 6 seconds will be 84ft not 126ft... ok.
so now, 120-70/6-3 = 16.7 ft/sec.

Answer 12

yes

Answer 13

no, thats wrong

Answer 14

how did you get y(t) = 86 feet at t = 6 seconds?

Answer 15

alright, looking at the y[t] im stuck.... now they got 41ft at the rate of 5 ft/s which means that was after 8.2 seconds... so after 6 seconds would be AFTER that 8.2 seconds....? so wouldn't i add them or wait, would it be 8.2+6 = 14.2 and then i use 14.2 x 5 =71 ft... so i'd use 71 ft then?

Answer 16

define when you start your timer. Suppose you had a timer to time these events. Tell me when you would start your timer.

Answer 17

x, y and s are functions of time, correct? So you have to start your time at some point. When do you start your time?

Answer 18

do you have an answer? or do you want me to explain?

Answer 19

at the 41ft, when the cycle and balloon are in a vertical line to each other... i'd start my timer then.

Answer 20

okay great!

Answer 21

lets define y(t) then. Can you give an expression for y(t)?

Answer 22

y[t] = 5t

Answer 23

so at t = 6 seconds, y(t) = 30 feet?

Answer 24

are you sure your definition is correct?

Answer 25

lets look at this another way. When you start your timer, how high is the balloon?

Answer 26

41ft.

Answer 27

so, at t = 0, y(t) = 41 ft

Answer 28

how high is the balloon 1 second after you started your timer?

Answer 29

46ft

Answer 30

how did you get 46 ft?

Answer 31

46 = 41ft+5ft/second*1second. correct?

Answer 32

so what is the general expression for y(t)?

Answer 33

y[t] = 5t+41

Answer 34

okay great!

Answer 35

so 6 seconds later, what is y(t)?

Answer 36

71ft

Answer 37

great!

Answer 38

have you done derivatives?

Answer 39

yes

Answer 40

this question is part of chapter on applications of derivatives.

Answer 41

okay.so you are asked to find ds(t)/dt at t = 3 seconds

Answer 42

lets define s(t). What is s(t)?

Answer 43

s[t] = square root of y[t]^2 plus x[t]^2

Answer 44

okay, so \[ds(t)/dt = d(y(t) ^{2}+x(t)^{2})^{0.5}/dt\]

Answer 45

right? and you know that dy/dt = 5 ft/sec and dx/dt = 14 ft/sec

Answer 46

solve the above expression in terms of dy/dt and dx/dt. Substitute for dy/dt and dx/dt and find the value of ds/dt at t = 3.

Answer 47

did you understand?

Answer 48

alright, i think i get it, lemme have a shot at it and ill post my answer in a bit.

Answer 49

note that dy/dt = d(41+5t)/dt = 5
and dx/dt = d(14t)/dt =14

Answer 50

note also that x(t) = 14t

Answer 51

and y(t) = 41+5t

Answer 52

alright, finding the derivative of teh first equation is:
0.5(y[t]^2 + x[t]^2)^-0.5 times (2y[t]+2x[t])
ok, using this differentiation, and knowing:
y[t]^2=3136 2y[t]=112
x[t]^2=1264 2x[t]=84
substituting it all in, i get the answer of 1.4 ft/sec.

Answer 53

okay great, I dont know what the answer is, but your differentiation seems correct.

Answer 54

alriiiight, thanks SO MUCH, you really were a GREAT help :] thanks ^^

Answer 55

you are welcome. You should do the other problem the same way too. the earlier one with the rectangle i mean.

Answer 56

good luck!