A bird flies from the bottom of the cave during the time period 0

Question

A bird flies from the bottom of the cave during the time period 0<orequalto t<orequalto 4 seconds. If the bird moves along a curve, so that its velocity at time t is v(t)= the square root of (t) -1, find the distance travelled by the bird in the time period.

v(t)= t√−1 m/s

anonymous · Answer

Is the equation: $$\sqrt{(t)-1}$$?

anonymous · Answer

$$v(t) = \frac{dx}{dt}$$

anonymous · Answer

$$dx = v(t) dt$$

anonymous · Answer

$$\int_0^x dx = \int_0^4v(t) dt$$

anonymous · Answer

no no sorry just the square root of t

anonymous · Answer

then minus one

anonymous · Answer

Now I am going to solve the RHS(Right Hand Side)

anonymous · Answer

$$v(t) = \sqrt t -1$$

anonymous · Answer

yes

anonymous · Answer

$$RHS = \int_0^4 \sqrt t dt-  dt$$

anonymous · Answer

$$RHS = \frac{2}{3} t^{3/2} - t |_0^4 $$

anonymous · Answer

Then On whole it becomes 

$$x = \frac{16}{3} - 4$$

anonymous · Answer

Its positive?

amistre64 · Answer

distance traveled would correlate to a line integral
$$\int\sqrt{[f'(x)]^2+[g'(x)]^2}$$

anonymous · Answer

i know it sounds stupid but my first time i got a negative number. I guess it was just another silly math mistake :)

anonymous · Answer

Ok I got Displacement Not Distance

anonymous · Answer

I don't know about the distance but displacement should be the one I got

anonymous · Answer

ohhh no.

anonymous · Answer

And Distance is always positive

amistre64 · Answer

since there is only 1 equation to deal with here; i believe it goes:

$$\int_{a}^{b}\  \sqrt{1+\left(\frac{1}{2\sqrt{t-1}}\right)^2}\ dt$$

anonymous · Answer

You can't have negative

anonymous · Answer

So if we have the displacement, how can we use that to find distance?

anonymous · Answer

Distance Traveled= $$\int\limits_{0}^{4} (\sqrt{(t)}-1) dx$$
=(t)^(1/2)-1
=(t)^(3/2)/(3/2)   -   (t)
=(2/3)(t^(3/2)) - (t)
Now plug in t=4 solve then t=0 solve.
Subtract the result of t=4 from result of t=0 

Distance=4/3

anonymous · Answer

Does that make sense?

anonymous · Answer

Thats just what I got justtheretohelp but it is Displacement

hero · Answer

The integral of velocity is distance, right?

hero · Answer

Oh, wait....

hero · Answer

Yeah, it's displacement

hero · Answer

lol, I agree

anonymous · Answer

Well the integral of the absolute value of the v(t) function from b to a would be distance

amistre64 · Answer

but then i believe you stated its sqrt(t) - 1

$$\int_{a}^{b}\  \sqrt{1+\left(\frac{1}{2\sqrt{t}}\right)^2}\ dt$$

$$\int_{a}^{b}\  \sqrt{1+\left(\frac{1}{4t}\right)}\ dt$$

$$\int_{a}^{b}\  \sqrt{\frac{4t+1}{4t}}\ dt$$

$$\int_{a}^{b}\  {\frac{\sqrt{4t+1}}{2\sqrt{t}}}\ dt$$

have a messed it up yet :)

anonymous · Answer

not to put in my two cents at the last minute, but what do you make of this line 

. If the bird moves along a curve, so that its velocity at time t is v(t)= the square root of (t) -1

anonymous · Answer

you are given the velocity, but not the curve!

amistre64 · Answer

...great, now we need to rescript the whole thing.  cut!!

hero · Answer

bravo!

anonymous · Answer

you have 
$$v(t)=\sqrt{t}-1$$  and so if the bird was moving in a straight line, then you would take the integral over positive and negative part of this and add, but we have no idea what the bird is doing do we?

anonymous · Answer

of course i could be wrong.

amistre64 · Answer

if theres an option to click "next"... take it now :)

anonymous · Answer

@amistre you are finding the length of the curve 
$$y=\sqrt{t}-1$$ from 0 to 4

amistre64 · Answer

that i was .... that i was

hero · Answer

lol

anonymous · Answer

now from 0 to 1 apparently the bird is flying backwards

anonymous · Answer

maybe it is a hummingbird in a cave.

anonymous · Answer

lol

anonymous · Answer

Gee.. I always thought distance and displacement were the same thing.. :/

anonymous · Answer

i will make a guess.  my guess is your teacher wants 
$$\int_0^1 1-\sqrt{t}dt +\int_1^4\sqrt{t}-1 dt$$

anonymous · Answer

no displacement is not the same as distance. if you go back 4 units and forward 6 your displacement is 2 and your distance is 10

anonymous · Answer

that is you traveled ten miles but are only two miles from home

anonymous · Answer

So how do we get to the distance

anonymous · Answer

integrate over where it is positive and negative separately.  then add

anonymous · Answer

Agree, with satellite73. Didn't think it through the first time.

anonymous · Answer

i should say integrate over where it is negative, then take absolute value, then add

amistre64 · Answer

could we integrate up to a position function?

anonymous · Answer

Gosh, and I thought mere integration was complex...

amistre64 · Answer

then the derivative of the position function would amount to the velocity function to use in the distance formula?

amistre64 · Answer

in other words, we already have h'(x) given right?

anonymous · Answer

$$\int_0^1\sqrt{t}-1dt=-\frac{1}{3},\int_1^4 \sqrt{t}-1 dt =\frac{5}{3}$$ so i will bet the answer is 
$$\frac{1}{3}+\frac{5}{3}=2$$

anonymous · Answer

that is my bet, but i am not placing odds

amistre64 · Answer

$$\int \sqrt{1+(\sqrt{t}-1)^2}\ dt$$

maybe?

anonymous · Answer

@amistre i think what this means is we have velocity, just want distance traveled.

anonymous · Answer

$$v(t)=S'(t)=\sqrt{t}-1$$ so 
$$S(t)=\frac{2}{3}t^{\frac{3}{2}}-t$$

amistre64 · Answer

right, and in my twisted brain im thinking that the v(t) given is the derivative of the position function to place inside of the line integral as [V'(t)]^2?

anonymous · Answer

you are thinking too hard. it was that "curve" that threw you off. threw you a curve me might say

amistre64 · Answer

:)  yousa mighta be right

amistre64 · Answer

if my thought is right, in some alternate universe as that may be :) ; I get this: http://www.wolframalpha.com/input/?i=integrate%28sqrt%28x-2sqrt%28x%29%2B2%29%29dx+from+0+to+4 abt 4.59 meters