anonymous
  • anonymous
ok, last question in the assignment that i can't seem to get, [derivative = hard ;_;], once again the question is attached in next post.
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
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anonymous
  • anonymous
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:
anonymous
  • anonymous
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

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anonymous
  • anonymous
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?
anonymous
  • anonymous
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
anonymous
  • anonymous
yes.
anonymous
  • anonymous
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.
anonymous
  • anonymous
soooo, is this correct? o:
anonymous
  • anonymous
how did you get x(t) = 126 ft for t =6sec?
anonymous
  • anonymous
um, at 3 sec the x[t] is = 42ft so at 6 sec its 14ft/sec times 6s = 84 ft + 42 ft = 126 ft
anonymous
  • anonymous
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?
anonymous
  • anonymous
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.
anonymous
  • anonymous
yes
anonymous
  • anonymous
no, thats wrong
anonymous
  • anonymous
how did you get y(t) = 86 feet at t = 6 seconds?
anonymous
  • anonymous
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?
anonymous
  • anonymous
define when you start your timer. Suppose you had a timer to time these events. Tell me when you would start your timer.
anonymous
  • anonymous
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?
anonymous
  • anonymous
do you have an answer? or do you want me to explain?
anonymous
  • anonymous
at the 41ft, when the cycle and balloon are in a vertical line to each other... i'd start my timer then.
anonymous
  • anonymous
okay great!
anonymous
  • anonymous
lets define y(t) then. Can you give an expression for y(t)?
anonymous
  • anonymous
y[t] = 5t
anonymous
  • anonymous
so at t = 6 seconds, y(t) = 30 feet?
anonymous
  • anonymous
are you sure your definition is correct?
anonymous
  • anonymous
lets look at this another way. When you start your timer, how high is the balloon?
anonymous
  • anonymous
41ft.
anonymous
  • anonymous
so, at t = 0, y(t) = 41 ft
anonymous
  • anonymous
how high is the balloon 1 second after you started your timer?
anonymous
  • anonymous
46ft
anonymous
  • anonymous
how did you get 46 ft?
anonymous
  • anonymous
46 = 41ft+5ft/second*1second. correct?
anonymous
  • anonymous
so what is the general expression for y(t)?
anonymous
  • anonymous
y[t] = 5t+41
anonymous
  • anonymous
okay great!
anonymous
  • anonymous
so 6 seconds later, what is y(t)?
anonymous
  • anonymous
71ft
anonymous
  • anonymous
great!
anonymous
  • anonymous
have you done derivatives?
anonymous
  • anonymous
yes
anonymous
  • anonymous
this question is part of chapter on applications of derivatives.
anonymous
  • anonymous
okay.so you are asked to find ds(t)/dt at t = 3 seconds
anonymous
  • anonymous
lets define s(t). What is s(t)?
anonymous
  • anonymous
s[t] = square root of y[t]^2 plus x[t]^2
anonymous
  • anonymous
okay, so \[ds(t)/dt = d(y(t) ^{2}+x(t)^{2})^{0.5}/dt\]
anonymous
  • anonymous
right? and you know that dy/dt = 5 ft/sec and dx/dt = 14 ft/sec
anonymous
  • anonymous
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.
anonymous
  • anonymous
did you understand?
anonymous
  • anonymous
alright, i think i get it, lemme have a shot at it and ill post my answer in a bit.
anonymous
  • anonymous
note that dy/dt = d(41+5t)/dt = 5 and dx/dt = d(14t)/dt =14
anonymous
  • anonymous
note also that x(t) = 14t
anonymous
  • anonymous
and y(t) = 41+5t
anonymous
  • anonymous
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.
anonymous
  • anonymous
okay great, I dont know what the answer is, but your differentiation seems correct.
anonymous
  • anonymous
alriiiight, thanks SO MUCH, you really were a GREAT help :] thanks ^^
anonymous
  • anonymous
you are welcome. You should do the other problem the same way too. the earlier one with the rectangle i mean.
anonymous
  • anonymous
good luck!

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