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April 19, 2014

April 19, 2014

Posted by **AP Physics** on Saturday, June 13, 2009 at 1:12pm.

Were given this eqaution

x = At^2 + B

were A = 2.10 m/s^2

B = 2.80 m

ok and this is an example problem and it's wlking us through it

find the instantaneous velocity at t=t2=5.00s equlas the slope of the tangent to the curve at point P2 whoen in Fig. 2-12b and we could measure this slope off the gropah to obtain v2. we can calculate v more precisely, and for any time using the give formula

x = At^2 + B,

which is the engiene's position x at time t. Using the calculus formula for derivitves (THIS STEP I'M LOST)

(d/(dt))(Ct^n)= nCt^(n-1) and (dC)/(dt) = 0,

where C is any given constant, then

v = (dx)/(dt) = (d/(dt))(At^2 + B) = 2At.

We are given A = 2.10 m/s^2, so for t=t2=5.00s,

v2 = 2At = 2(2.10 m/s^2)5.00s = 21.0 m/s

ok it has been a long time sense I have taken calculus. I understand why instanteous velociy is equal to (dx)/(dt) but I do not understand this problem. Can you please explain it to me. I got lost were I wrote (THIS STEP I'M LOST)

I don't understand this

"(d/(dt))(Ct^n)= nCt^(n-1) and (dC)/(dt) = 0,

where C is any given constant, then

v = (dx)/(dt) = (d/(dt))(At^2 + B) = 2At."

Please explain to me the constant thing and what d is all by itself and what is dC and why they divide it by dt and how they got 2At

Thank you!

- AP Physics -
**Anonymous**, Saturday, June 13, 2009 at 3:33pmIn general

if f(x) = x^n

then d f(x)/dx = n x^(n-1)

so

if f(x) = x^2

then d f(x)/dx = 2 x

so if f(t) = t^2

df(t)/dt = 2 t

the constant multiple remains (three times the function has three times the slope) .

if f(t) = 2.1 t^2

then

d f(t)/dt = 4.2 t

Now about that constant B

The slope of a constant is zero.

so

d/dt (A t^2 + B)

= 2 A t + 0

the end except to prove that d/dx (x^n) = n x^(n-1)

f(x+dx) = (x+dx)^n

binomial expansion

(x+dx)^n = x^n + C(n,1) x^(n-1) dx + C(n,2) x^(n-2) (dx)^2 ....... +dx^n

C(n,k) = binomial coefficient

= n!/[(n-k!) k!]

so for Cn,1)

C(n,1) = n!/[(n-1)!] = n (lo and behold)

so

f(x+dx) = x^n + n x^(n-1)dx + terms in dx^2 through dx^n

so

f(x+dx) -f(x) = n x^(n-1) dx + terms in dx^2 to dx^n

so

df(x)/dx = [f(x+dx)-f(x)]/dx

as dx ---> 0

= n x^(n-1)

f(x) = x^n

f

- AP Physics -
**Damon**, Saturday, June 13, 2009 at 3:37pmIgnore the last two lines, I was thinking out loud.

You may not need that proof at the end that d x^n/dx = n x^(n-1) but it is not a bad idea to know where the derivative comes from.

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