Question

How do i solve this initial value problem:

Answer 1

and where's the problem?

Answer 2

\[ds/dt=8\sin^2(t+(\pi/2)), s(0)=8\]

Answer 3

can someone just tell, what it is we are trying to do here in this problem

Answer 4

are we integrating to find F or what? i dont get

Answer 5

\[ds/dt=8\sin^2(t+(\pi/2))\]

\[ds=8\sin^2(t+(\pi/2))dt\]

\[\int ds=\int8\sin^2(t+(\pi/2))dt\]

\[=\cdots\]

Answer 6

now, when integratins ds, does that just intergrate to s

Answer 7

yes \[\int ds=s+c\]

Answer 8

can i ask you one more thing...

Answer 9

I had this problem :
\[6+\int\limits_{a}^{x}f(t)/t^2 dt=2\sqrt{x}\]

Answer 10

i was told : for all x>0 find the function f and a number a such that

Answer 11

if you need to find f...then use the fundamental theorem of calculus and differentiate both sides..then solve for f

Answer 12

are we trying to isolate f(t)

Answer 13

f(x)

Answer 14

now what about this number "a", what is taht suppose to be?

Answer 15

\[\frac{d}{dx}\int\limits_{a}^{g(x)}f(t)dt=f(g(x))g'(x)\]

Answer 16

oh i see a is suppose to be like the lower limit of integration of something like that

Answer 17

yes...so it disappears once you differentiate.

Answer 18

how would i go about solving for that?

Answer 19

you need to differentiate both sides of your equation

Answer 20

i got : f(x)=x^2/x^1/2

Answer 21

oh then just plug a into that

Answer 22

the a plays no role in this problem

Answer 23

okay, but i was told to find a number "a"

Answer 24

find a number a such that what ?

Answer 25

such that:

\[6+\int\limits_{a}^{x}f(t)/t^2dt=2\sqrt{x}\]

Answer 26

ok...jas

Answer 27

what?

Answer 28

jas...just a sec

Answer 29

okay

Answer 30

look like 9

Answer 31

replace \[f(x)=x^{3/2}\] into the equation then integrate

Answer 32

what r u saying? like 6+x^3/2=2sqrtx

Answer 33

\[6+\int\limits_{a}^{x}\frac{t^{3/2}}{t^2} dt=2\sqrt{x}\]

Answer 34

\[6+\int\limits_{a}^{x}t^{-1/2} dt=2\sqrt{x}\]

Answer 35

\[6+\left.2t^{1/2}\right|_a^{x}=2\sqrt{x}\]

\[6+2x^{1/2}-2a^{1/2}=2\sqrt{x}\]


\[6+2\sqrt{x}-2\sqrt{a}=2\sqrt{x}\]

Answer 36

a=9

Answer 37

no more questions from me after this one, why is it that we find the "a" this way?

Answer 38

how else would we find it?

Answer 39

but, i mean, why is this the way that we find it?

Answer 40

we need to isolate a. if a is stuck on the integral sign I can't isolate it.

Answer 41

thanks Zarkon, you the best, say hello to buzz for me

Answer 42

you know from toystory

Answer 43

Zarkon is from Voltron ;)