Question

I know the initial position is zero, and the final position is 32, but i can't find the total distance traveled for the life of me. Please help?!?!?!
http://webwork.math.ttu.edu/wwtmp/equations/7c/e361012b1a74824e9bc9c6d10bb4ed1.png

Answer 1

the integration of a curve tells us the displacement as well

Answer 2

what do you mean?

Answer 3

i mean that when we integrate a function along an interval we are finding the distance traveled

Answer 4

i have a hard time flipping back and forth; can you type that thing up in here so I can keep an eye on it :)

Answer 5

I'm in cal 1 and we have not discussed integrals.. so, I don't think this is the way my professor expects us to solve the answer. and sure give me one second

Answer 6

\[\S(t)=t^4-4t^3+8t for 0\let \le4\]

Answer 7

oops forget that for0

Answer 8

define what you mean by distance traveled then.

Answer 9

i=t should have been\[0\le t \le4\]

Answer 10

either we need to measure the curve itself; or we want to know how far we have moved from our starting point.

Answer 11

ok im not sure how to use the greater than sign obviously it should be [0,4], and I guess its measuring the curve itself because the answer is not 32

Answer 12

s(t) = t^4 -4t^3 + 8t ; from s(t) = 0 to s(t) = 32

Answer 13

ok, but measureing the curve is just as much integration as displacement.

you might be calling it antiderivatives tho

Answer 14

hmmm, i don't think we've talked about that either. oh well, how do you do it haha

Answer 15

its based upon measuring the distance between 2 points along the curve: we are rectifying the curve, or straightening it out

Answer 16

ok that makes sense

Answer 17

how would we find /\s ?

Answer 18

think pythag thrm

Answer 19

is that delta 5 or delta s

Answer 20

delta5 would be a bit of an obfuscation; so lets try delta s :)

Answer 21

"s"egment is the usual textbook term

Answer 22

lol ok well I'm guess in deltax^2 + deltay^2 = sort(delta5 lol s)

Answer 23

sort should have been sqrt

Answer 24

correct :)

\[\Delta s^2=\Delta x^2+\Delta y^2\]
\[\Delta s=\sqrt{\Delta x^2+\Delta y^2}\]
\[\sum\Delta s=\sum \sqrt{\Delta x^2+\Delta y^2}\]

this make sens so far?

Answer 25

we are adding up all the segment peices to get the total length

Answer 26

ok. so far so good

Answer 27

now, as we go infinitesimal here, our deltas disappear and we are left looking at the ghost of departed values; ds, dx, and dy:

\[\sum\Delta s=\sum \sqrt{\Delta x^2+\Delta y^2}\ \to \int_C ds=\int^{b}_{a}\sqrt{dx^2+dy^2}\]

Answer 28

the rate of change in x with respect to x is dx/dx ... or simply 1

the rate of change in y with respect to x is dy/dx; y'

Answer 29

\[\int^{b}_{a}\sqrt{1+(y')^2}\]

Answer 30

so where do we go from there or is that our answer?

Answer 31

we insert our interval into a and b; and insert our y' into its cozy little spot; then take the antiderivative (the integral) and see if we come up with the answer in the book to determine if we are even doing this right :)

Answer 32

s(t) = t^4 -4t^3 + 8t

s' = 4t^3 -12t^2 +8

and since the variable are just names, we can just apply the appropriate notations

\[\int^{b}_{a}\sqrt{1+(y')^2}dx\ \to \int^{b}_{a}\sqrt{1+(4t^3 -12t^2 +8)^2}dt\]

Answer 33

and you say this is from 0 to 4 correct?

Answer 34

http://www.wolframalpha.com/input/?i=integrate+%28sqrt%281%2B%284t%5E3+-12t%5E2+%2B8%29%5E2%29%29+from+0+to+4

thats about 50 if we did it right

Answer 35

yes 0 to 4

Answer 36

dude your a beast!!! that was right!

Answer 37

cool :)