Question

A hawk flying at 11 m/s at an altitude of 132 m accidentally drops its prey. The parabolic trajectory of the falling prey is described by the equation
y = 132 − (x^2/33)
until it hits the ground, where y is its height above the ground and x is its horizontal distance traveled in meters. Calculate the distance traveled by the prey from the time it is dropped until the time it hits the ground. Express your answer correct to the nearest tenth of a meter.

http://i1084.photobucket.com/albums/j409/QRAWarrior/MATA36-IMG009.png

Answer 1

@amistre64
(Ok amistre, here is my try)

Answer 2

Apparently, this was marked "wrong" by the e-assignment system.

Answer 3

The assignment said to round to the nearest tenth. I am pretty sure I did that.

Answer 4

lets see how we can accomodate it; also, now that i see the graph, you mighta measured this with respect to y a bit simpler eh

Answer 5

Sorry if nearest tenth meant 125.5, I tried that too just now and I got it wrong..

Answer 6

y = 132 − (x^2/33)

x^2 = 33(123 - y)

x = sqrt(33(123 - y))

x = sqrt(4059 - 33y)

\[\int_{0}^{132}\sqrt{1+(\frac{-33 }{2\sqrt{4059 - 33y}})^2}\ dy\]

Answer 7

prolly not easier is it :)

Answer 8

y = 132 − (x^2/33)

y' = -2x/33

33 tan(u) = -2x

33sec^2(u) du/-2 = dx \[-\frac{33}{2}\int_{}^{}sec^3(u) du\]

Answer 9

That is a secant cubed right?

Answer 10

yes

Answer 11

Ok good.

Answer 12

we can integrate that by tables :)

Answer 13

i had it all latexed up but the aw snap attack got it instead

Answer 14

@amistre64 you there? I think I already integrated the sec^3(x) because I got it from some previously done problem...

Answer 15

i see it, and its good now if you wanna keep the trig, youre gonna have to check those integration limits to be sure they are good

Answer 16

u = arctan(-2x/33) at x=0 and x=88 right?

Answer 17

x=0 and x=66

Answer 18

so 0 and -1.3258, or keep that one in arctan mode till the final

Answer 19

Ok

Answer 20

OS is doing alot of sitting there and doing nothing tonight :/

Answer 21

OS?

Answer 22

@amistre64

Answer 23

@amistre64
@TuringTest
@myininaya
@SmoothMath
@shivam_bhalla
@Zarkon
@dumbcow
@satellite73
@AccessDenied
@apoorvk

Help anyone? We almost got this problem except near the end...

Answer 24

Hmm, I guess we just complicated the situation much more than it is. Let's refresh.

Answer 25

So, 'y' is the height of the falling prey, and it has been expressed in terms of 'x' which is the distance it travels simultaneously in the horizontal direction as it falls. right?

Answer 26

there?

Answer 27

@apoorvk

Answer 28

Yeah so clear that part?

Answer 29

Yes

Answer 30

But we need to find the arclength of the distance it travelled.

Answer 31

Ohkay, misread the question. Let me think now.

Answer 32

I have a suggestion: maybe there is somewhere I messed up in my original attempt (first post I believe)..

Answer 33

I am thinking - that the integration of the equation would actually end up giving the area covered under the arc during the drop - what do you think?

Answer 34

Actually, it gives the length of the curve, as if you took it off of the 2D-plane, straightened it, and measured it with a ruler

Answer 35

@FoolForMath ?

Answer 36

@satellite73
@.Sam.
@Callisto
@ash2326
@blues

?

Answer 37

Anyone who just came: please look at the opening post.

Answer 38

I have a full attempt there.

Answer 39

@mahmit2012

Answer 40

@jamesj

Answer 41

Uh everybody is able to VIEW the attempt clearly right? If not, let me know.

Answer 42

@satellite73