Question

The position of a particle at time t is s(t)=(t^3) +t . How do I ESTIMATE the instantaneous velocity at t=1 without using a calculator?

Answer 1

Well what do you know about the relationship[ between position and velocity?

Answer 2

velocity is the derivative of the position?

Answer 3

yes so you find

s'(t) = v(t)

thgen evaluate v(t) at t=1

so v(1) = ?

Answer 4

oh, i'm not supposed to find the derivative because my teacher wants us to use: as the limit approaches...

Answer 5

i mean, both methods are finding the derivative, but i'm not supposed to "know the easy way (your way)" yet

Answer 6

ah so she wants you to use the definition of the derivative meaning....

\[f'(x) = \lim_{h \rightarrow 0} \frac{ f(x+h) - f(x) }{ h }\]

Answer 7

yes, like that, except for the minor change of changing the 'f' to 's' and 'h' to 't'

Answer 8

well x becomes t but f does become s yes so

Answer 9

if i want the instantanoues velocity at t=1, should i write as t ->0 or as t->1 under 'lim'?

Answer 10

nah the h stays the x is what becomes the t

Answer 11

so this is what you are after

\[v(t) = \lim_{h \rightarrow 0}\frac{ s(t+h) - s(t) }{ h }\]

Answer 12

what does the 'h' refer to?

Answer 13

it is basically the change in the variable so some textbooks refer to the h as this....

\[f'(x) = \lim_{\Delta x \rightarrow 0} \frac{ f(x+\Delta x )-f(x)}{ \Delta x }\]

Answer 14

so it's always delta (variable) ->0 and not approaching another value, like 1?

Answer 15

\[
v(1) = \lim_{\Delta t \rightarrow 0}\frac{ s(1+\Delta t) - s(1) }{ \Delta t }
\]

Answer 16

no it approaches 0 because if you were to use the formula the 0 makes all the change in disappear and you are left with a formula that you can plug your value in or you can shortcut it the way aum has by putting the 1 in right way

Answer 17

um what do i do if i want an estimate of v(1)?

Answer 18

sorry

Answer 19

your estimate is when you plug the value in. If it is asking for a estimate without putting a value in you can't really to be honest

Answer 20

\[
v(1) = \lim_{h \rightarrow 0}\frac{ s(1+h) - s(1) }{ h } =
\lim_{h \rightarrow 0}\frac{ (1+h)^3 + (1+h) - (1^3 + 1) }{ h } = ?
\]

Answer 21

yea that will be your answer

Answer 22

\( (1+h)^3 = 1 + 3h + 3h^2 + h^3\)

Answer 23

in the end i got 4+4h. is that correct?

Answer 24

make the h become 0 you get 4

Answer 25

\[

v(1) = \lim_{h \rightarrow 0} 4 + 4h = 4\]

Answer 26

oh yeah! i forgot to plug in the 0!
and the 4 here is the same as 4 i would have gotten if i used the shorter method, i.e. 3t squared + 1

Answer 27

Yes. v(t) = 3t^2 + 1
v(1) = 3 + 1 = 4

Answer 28

So yours is not really an estimate but an accurate value done through limits instead of using derivatives formula.

Answer 29

As I said i wouldn't know how to do it as an estimate

Answer 30

my textbook says "To estimate the instantaneous rate of change at x = x0, compute the average rate of change over several intervals[x0,x1](or[x1,x0]) for x1 close to x0. "

Answer 31

ah they want you to do that?

Answer 32

In that case, v(1) = ( s(2) - s(0) ) / ( 2 - 0 )

Answer 33

and to find the estimated instantaenous velocity, "compute the average rate for several intervals lying to the left and right of" t=1

Answer 34

hilbert, i think so - but i would need a calculator for the decimals (like t=1.0001 and t= .9999), but the teacher says i can't use a calculator

Answer 35

I find this question flawed to be honest

Answer 36

\[
v(1) \approx \frac { s(2) - s(0) }{ 2 - 0 } = \frac{2^3 + 2 - 0^3 - 0}{2-0} =
\frac{10}{2} = 5

\]

Answer 37

aum, i'm curious about why it's s(2)?

Answer 38

he is taking a number to the right of 1

Answer 39

one unit to the left of t = 1 and one unit to the right of t = 1
t = 0 and t = 2

Answer 40

To get a better estimate we can go 0.5 on either side but you have to calculate (0.5)^3 and (1.5)^3 if you want to do that.

Answer 41

i see. but yea, id rather use whole numbers since i'm not allowed to use a calculator

Answer 42

Actually 0.1 on either side may not be that hard to do.

Answer 43

should there be another equation of v(1) in addition to the one you posted earlier, aum?

Answer 44

oh

Answer 45

\[
v(1) \approx \frac { s(1.1) - s(0.9) }{ 1.1 - 0.9 } =
\frac { 1.1^3 + 1.1 - 0.9^3 - 0.9 }{ 0.2} = \\
\text{ } \\
\frac { 1.331 + 1.1 - 0.729 - 0.9 }{ 0.2} =
\frac{0.802}{0.2} = \frac{0.802}{0.2} * \frac{5}{5} = \frac{4.01}{1} = 4.01



\]

Answer 46

thank you to both; you taught me stuff i didn't know

Answer 47

You are welcome.