Instantaneous Velocity

[Jason76]

How would you find the instantaneous velocity at \(\displaystyle x = 1\) for \(\displaystyle y = 15t - 1.86t^{2}\) IF you did NOT know what a derivative is?

For example, the professor would need work to show the answer, so taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would not be possible. Of course, taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would get you the answer, assuming you used that method.

 

[HallsofIvy]

How would you find the instantaneous velocity at \(\displaystyle x = 1\) for \(\displaystyle y = 15t - 1.86t^{2}\) IF you did NOT know what a derivative is?

For example, the professor would need work to show the answer, so taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would not be possible. Of course, taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would get you the answer, assuming you used that method.

First, don't use "y" for the function in one place and "f" in another. Also, do not use "t" in the function and then talk about the derivative at "x= 1".

If you are not allowed to use the derivative, then use the definition of the derivative: \(\displaystyle \lim_{h\to 0}\frac{y(1+ h)- f(1)}{h}\). Of course, you can't actually find the "instantaneous" velocity- that would, pretty much by definition, be the derivative. Instead calculate that "difference quotient" for very small "h"
to get an average velocity over a small interval.

 

[Deleted member 4993]

How would you find the instantaneous velocity at \(\displaystyle x = 1\) for \(\displaystyle y = 15t - 1.86t^{2}\) IF you did NOT know what a derivative is?

For example, the professor would need work to show the answer, so taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would not be possible. Of course, taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would get you the answer, assuming you used that method.

The measure of instantaneous velocity - by definition - is the first time-derivative of the position vector.

I do not know - why you should NOT know what the first derivative is and still be asked to calculate the instantaneous velocity!!?!!

 

[DrPhil]

How would you find the instantaneous velocity at \(\displaystyle x = 1\) for \(\displaystyle y = 15t - 1.86t^{2}\) IF you did NOT know what a derivative is?

For example, the professor would need work to show the answer, so taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would not be possible. Of course, taking the \(\displaystyle f '(1)\) of \(\displaystyle y = 15t - 1.86t^{2}\) would get you the answer, assuming you used that method.

When I first took physics in high school, I had never heard of calculus [no such thing in H.S. way back then!!]. So we memorized the equation of motion:

\(\displaystyle \displaystyle y(t) = y_0 + v_0t + (1/2)a\ t^2\)

Thus I knew that the number multiplying \(\displaystyle t\) would be a velocity. We also could see from the units that velocity as a function of \(\displaystyle t\) would be

\(\displaystyle \displaystyle v(t) = v_0 + a\ t\)

 

[Deleted member 4993]

When I first took physics in high school, I had never heard of calculus [no such thing in H.S. way back then!!]. So we memorized the equation of motion:

\(\displaystyle \displaystyle y(t) = y_0 + v_0t + (1/2)a\ t^2\)

Thus I knew that the number multiplying \(\displaystyle t\) would be a velocity. We also could see from the units that velocity as a function of \(\displaystyle t\) would be

\(\displaystyle \displaystyle v(t) = v_0 + a\ t\)

But the above is true for constant acceleration - Galileo did not know differentiation either.

 

[Jason76]

First, don't use "y" for the function in one place and "f" in another. Also, do not use "t" in the function and then talk about the derivative at "x= 1".

If you are not allowed to use the derivative, then use the definition of the derivative: \(\displaystyle \lim_{h\to 0}\frac{y(1+ h)- f(1)}{h}\). Of course, you can't actually find the "instantaneous" velocity- that would, pretty much by definition, be the derivative. Instead calculate that "difference quotient" for very small "h"
to get an average velocity over a small interval.

\(\displaystyle \lim x \rightarrow 0 \dfrac{f(x+ h) - f(x)}{h} \)


\(\displaystyle \lim x \rightarrow 0 \dfrac{f(15t - 1.86t^{2} + h) - f(15t - 1.86t^{2})}{h} \)

 

[Deleted member 4993]

\(\displaystyle \lim x \rightarrow 0 \dfrac{f(x+ h) - f(x)}{h} \)


\(\displaystyle \lim x \rightarrow 0 \dfrac{f(15t - 1.86t^{2} + h) - f(15t - 1.86t^{2})}{h} \).....................INCORRECT

\(\displaystyle \displaystyle \lim_{x \to 0} \dfrac{[15(t+h) - 1.86(t+h)^{2} + h] - (15t - 1.86t^{2})}{h} \)

 

[Bob Brown MSEE]

Instead calculate that "difference quotient" for very small "h"
to get an average velocity over a small interval.

1) If y(t) and y(t+h) are known exactly.
2) If you have constant acceleration.
3) If you calculate that "difference quotient" for very small "h".
4) you get the average velocity over a small interval h.

However, you have the EXACT velocity at the midpoint t+h/2.

EDIT: btw, h does not have to be "very small".

 

[DrPhil]

1) If y(t) and y(t+h) are known exactly.
2) If you have constant acceleration.
3) If you calculate that "difference quotient" for very small "h".
4) you get the average velocity over a small interval h.

However, you have the EXACT velocity at the midpoint t+h/2.

The EXACT velocity is

......\(\displaystyle v(t) = v_0 + a\ t = 15 - 3.72\ t\)

 

[Bob Brown MSEE]

My Solution

The EXACT velocity is

......\(\displaystyle v(t) = v_0 + a\ t = 15 - 3.72\ t\)

My Solution (not using derivative) using average velocity.
Given: y(t) = 15t -1.86tt
This meets the criteria that I listed (above).
Also y(0)=0
so finding average v over (0, 2t) implies that v(t) = y(2t)/2
ANSWER: v(1) = y(2)/2 = 15 - 1.86*2 = 11.28

Check using Dr Phil's formula.
v(t) = v₀+ at = 15-3.72t
v(1) = 11.28

Teacher Wants you to use the formulas for v(t) and y(t), most likely.
Dr Phil's approach is likely what your teacher wants.

Learn
However, learning that ...
average velocity = EXACT velocity, ( for constant acceleration )
can save you time and allow you to do many problems in your head.

Even more importantly (since this is the Calculus Forum)...
The definition of derivative...
\(\displaystyle \lim h \rightarrow 0 \dfrac{f(x+ h) - f(x)}{h} \)
is very useful in proofs,
and also useful as a formula to numerically get an approximate value of a derivative.

This definition of derivative...
\(\displaystyle \lim h \rightarrow 0 \dfrac{f(x+ h) - f(x-h)}{2h} \)
is much more useful as a formula to numerically get an approximate value of a derivative.

The first limit is a linear approximation, but the second limit is a quadratic approximation. For analytic functions over (x-h, x+h) you will get a MUCH better numerical approximation using the second limit.

As we see in this post, for quadratic functions the second limit is EXACT for ALL h.