So the equation given is x=20+5t^2
x= distance and t=time. I am supposed to find the instantaneous velocity using the equaiton v=dx/dt
I know that d is for derivative, but I have not studied derivatives yet.
The textbook also noted that for any n, the derivative of t^n is nt^(n-1). I suppose this means that the derivative of t^2 is 2t? but wouldn't the 2t's cross out if placed within the equation? Is d in dx different from d in dt? I am obviously missing something. Could someone please explain this to a student who has not begun precalc?
Thank you for your time.
x= distance and t=time. I am supposed to find the instantaneous velocity using the equaiton v=dx/dt
I know that d is for derivative, but I have not studied derivatives yet.
The textbook also noted that for any n, the derivative of t^n is nt^(n-1). I suppose this means that the derivative of t^2 is 2t? but wouldn't the 2t's cross out if placed within the equation? Is d in dx different from d in dt? I am obviously missing something. Could someone please explain this to a student who has not begun precalc?
Thank you for your time.
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Yes, what you're missing is that the "d" in dx/dt is not a variable. You have to read the whole thing as one unit. Or you can read "d/dt" as one thing, an operation called "differentiate with respect to time" and it's operating on x.
It's a notation for a particular kind of operation. And when you apply the "d/dt" operation on the function (20 + 5t^2) it produces the function 10t.
That's because it produces 5*df/dt from 5f, and it produces 0 from the constant 20 and 2t from t^2. So d/dt[20 + 5t^2] = 0 + 5*(2t) = 10t.
It's a notation for a particular kind of operation. And when you apply the "d/dt" operation on the function (20 + 5t^2) it produces the function 10t.
That's because it produces 5*df/dt from 5f, and it produces 0 from the constant 20 and 2t from t^2. So d/dt[20 + 5t^2] = 0 + 5*(2t) = 10t.
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The instantaneous velocity for this equation is the first derivative.
v = 10t
now for any time you can find the instantaneous by plugging in for t.
v = 10t
now for any time you can find the instantaneous by plugging in for t.
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No you are substituting 2t for t²
You also need to know that the derivative of a constant is zero
so dx/dt = 5(2t) = 10t
This is a poor question if you don't have calculus.
You also need to know that the derivative of a constant is zero
so dx/dt = 5(2t) = 10t
This is a poor question if you don't have calculus.
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v = dx/dt = 10t
This gives the velocity at any specific time.
Reading up on 'differentiating polynomials' will help further.
:)>
This gives the velocity at any specific time.
Reading up on 'differentiating polynomials' will help further.
:)>
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You can think of dx/dt as a very small ratio of delta x and delta t -> this means that for a very small amount of change in t (this value being dt), a very small change in x occurs (this value is dx).
Using the familiar equation speed = distance / time, you can see that the instantaneous velocity is equal to the very small dx divided by the very small dt:
v = dx / dt
As for the derivative itself, you have the rule correct. If x = 20 + 5t^2, then dx / dt = 10*t
The dt in the left side of the equation is not a true value, which is why you cannot put it on the right side of the equation. dx/dt is simply a way of denoting "the derivative of x with respect to t".
Using the familiar equation speed = distance / time, you can see that the instantaneous velocity is equal to the very small dx divided by the very small dt:
v = dx / dt
As for the derivative itself, you have the rule correct. If x = 20 + 5t^2, then dx / dt = 10*t
The dt in the left side of the equation is not a true value, which is why you cannot put it on the right side of the equation. dx/dt is simply a way of denoting "the derivative of x with respect to t".