Question

F(x) = The integral from 15 to x of sqrt(t^2 + 64) dt. Evaluate at F(15) and F'(15)

Answer 1

\[\int\limits_{15}^{x} \sqrt{t ^{2}+64} dt\]

Answer 2

\[F(15)=\int _{15}^{15}\sqrt{t^2+64}dt=0\] because the "length of the path" is zero

Answer 3

it is always the case that
\[\int_a^af(t)dt=0\]

Answer 4

Ok, second part: I'm not that great with integrals and I'm unsure how to integrate the function. Using u-sub for t^2 + 64 would get du of 2t. But since there is no t outside the function, I'm not quite sure what to do...

Answer 5

you are not asked to integrate the function
don't worry about finding a nice closed form for the integral
you are only asked for \(F(15)\) and \(F'(15)\)

Answer 6

it is a common misunderstanding that given some function \(f(tx)=\text{whatever}\) then somehow one can find some other function in a nice closed form \(F(x)\) with \(F'(x)=f(x)\)

Answer 7

it is almost never the case

Answer 8

Ok. So I get that F(15) = 0. You can't have a length or area under a distance of nothing. However, what does it mean when it asks F'(15) ? What should I be looking to do at that point?

Answer 9

the derivative of the integral is the integrand

Answer 10

So it should be \[\sqrt{t^{2}+64}\], yeah? At least, I already tried that but the program says it is incorrect

Answer 11

well actually it should be
\[\sqrt{x^2+64}\] since \[F(x)=\int _{15}^x\sqrt{t^2+64}dt\] is a function of \(x\) not \(t\)
so \(F'(x)=\sqrt{x^2+64}\)

Answer 12

But it's F'(15)

Answer 13

you were asked not for \(F'(x)\) but rather \(F'(15)\) right?

Answer 14

Yes. I get that x replaces t when it is asked as a function of x. Just not quite sure when its 15

Answer 15

i think you got confused maybe because you were thinking of integrals
suppose you were in a basic pre - calc course and were given the problem as follows :

if \(f(x)=\sqrt{x^2+64}\) what is \(f(15)\)?

i bet you could solve in seconds

Answer 16

Bleh. 17.

Answer 17

lol

Answer 18

I was. One last one. so if it says F"(15). Take the derivative of f(x) and evaluate at 15?

Answer 19

i guess

Answer 20

lol, thanks. If you have a moment, may I ask another question?

Answer 21

\[\frac{x}{\sqrt{x^2+64}}\] is what i get

Answer 22

i mean as the second derivative
at 15 i guess you get \(\frac{15}{17}\)

Answer 23

I got the same

Answer 24

yeah go ahead and ask

Answer 25

I'm working on finding a total distance traveled for rectilinear motion.
\[v(t) = t ^{3} - 3t ^{2} + 2t; 0\le t \le 5\]

Answer 26

I get \[s(t) = t ^{4}/4 -t ^{3} + t ^{2}\]

Answer 27

And I got displacement as equal to 225/4 which worked out. But I'm unsure how to find total distance traveled

Answer 28

be careful here because my guess is you have to break up the integral in to parts where it is going forward and going backwards

Answer 29

For the equation I found that t = 2 is where it goes from negative to positive. However, I don't know what to do with that information

Answer 30

integrate from 0 to 2, then from 2 to 5
first one is positive, second one is negative, then add the absolute values
i did not check that 2 was right, but i believe you

Answer 31

But if 2 is zero and 0 is zero, then the integral from 0 to 2 is just zero?

Answer 32

Then 2 to 5 is the same as 0 to 5 because its still 5 - 0

Answer 33

hold on one second, let me check this

Answer 34

I think this is where I'm getting pretty confused

Answer 35

ok first off, it is
\[f(t)=t^3-3t^2+2t\] right?

Answer 36

Yes

Answer 37

which by some happy coincidence is \(f(t)=t(t-1)(t-2)\)

Answer 38

so in fact unfortunately for you it changes sign at \(1\) and at \(2\)

Answer 39

Evaluate at the base function not at the integral

Answer 40

on the interval \([0,5]\) it is positive, then negative, then positive

Answer 41

First problem found

Answer 42

yeah hold on
it is the "speed" except it is actually "velocity" which can be negative
you are in your car driving forward for one hour, then backwards for one hour, then forwards for 3 hours

your "displacement" is just how far you are from home, but the total distance travelled is the forwards distance plus the backwards distance plus the forwards distance

Answer 43

So I get that from 0 to 1, it is 1/4, 1 to 2 it is 1/4 again(taking abs value) and then my original value of 225/4 from 2 to 5. Does this seem about right?

Answer 44

well i didn't to it, but that is certainly the right idea

Answer 45

Yeah, I just didn't see how I was messing up because I knew I needed separate values to add together showing total forward/backward distance

Answer 46

Perfect, thanks again!

Answer 47

yw