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!
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
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.
Certainly! I'll explain it step by step.
In calculus, the derivative of a function measures the rate at which the function is changing at a specific point. In this case, we're interested in finding the instantaneous velocity, which is the rate at which the position (x) is changing with respect to time (t).
To find the derivative of a function, we use the formula:
(d/dt)(Ct^n) = nCt^(n-1),
where C is any given constant, n is the exponent, and (d/dt) represents the derivative with respect to time.
In the given problem, the position function is given as x = At^2 + B, where A and B are given constants. To find the derivative of this function, we consider A as a constant and differentiate each term separately. Since B is constant, its derivative is zero ((dC)/(dt) = 0).
To find the derivative of At^2, we apply the derivative formula:
(d/dt)(At^2) = 2At^(2-1) = 2At.
So the derivative of x with respect to t, (dx)/(dt), is equal to 2At. This represents the instantaneous velocity at any given time.
Now, let's calculate the instantaneous velocity at t = t2 = 5.00s. We are given A = 2.10 m/s^2. Substituting the values into the equation:
v2 = 2At = 2(2.10 m/s^2)(5.00s) = 21.0 m/s.
So, at t = 5.00s, the instantaneous velocity, v2, is 21.0 m/s.
I hope this explanation clarifies the steps involved and answers your questions. Let me know if you need further clarification!