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AP Physics

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can you please explain this to me...

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 - ,

    In 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 - ,

    Ignore 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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