Approximating displacement The velocity in ft / s of an object moving along a line is given by v

Approximating displacement The velocity in ft / s of an...

Approximating displacement The velocity in $\mathrm{ft} / \mathrm{s}$ of an object moving along a line is given by $v=\sqrt{10 t}$ on the interval $1 \leq t \leq 7,$ where $t$ is measured in seconds. a. Divide the interval [1,7] into $n=3$ subintervals, [1,3] $[3,5],$ and $[5,7] .$ On each subinterval, assume the object moves at a constant velocity equal to $v$ evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [1,7] (see part (a) of the figure). b. Repeat part (a) for $n=6$ subintervals (see part (b) of the figure). (Check your book to see figure)

Approximating displacement The velocity in $\mathrm{ft} / \mathrm{s}$ of an object moving along a line is given by $v=\sqrt{10 t}$ on the interval $1 \leq t \leq 7,$ where $t$ is measured in seconds.
a. Divide the interval [1,7] into $n=3$ subintervals, [1,3] $[3,5],$ and $[5,7] .$ On each subinterval, assume the object moves at a constant velocity equal to $v$ evaluated at the midpoint of the subinterval, and use these approximations to estimate the displacement of the object on [1,7] (see part (a) of the figure).
b. Repeat part (a) for $n=6$ subintervals (see part (b) of the figure).
(Check your book to see figure)

Key Concepts

Midpoint Rule

The midpoint rule is a variant of the Riemann sum where the function is evaluated at the midpoint of each subinterval. This rule often offers a better approximation than using endpoints because it tends to balance the error over each interval, making it a common technique in numerical integration methods.

Riemann Sums

Riemann sums provide a method to approximate the value of a definite integral by partitioning the interval into smaller subintervals, evaluating the function at selected points within these subintervals, and summing the products of these values with the subinterval widths. This approximates the area under the curve, which in physical applications corresponds to quantities like displacement.

Velocity and Displacement Relationship

In kinematics, displacement over an interval is obtained by integrating the velocity function with respect to time. When the velocity is provided as a function of time, understanding this relationship is crucial because it connects the motion (through velocity) to the overall movement (displacement) of an object.

Definite Integral as Accumulated Change

The process of integrating a function over an interval represents the accumulation of small changes, such as the displacement accumulated over successive time intervals. This is foundational in calculus and applied scenarios where continuous changes are approximated by summing contributions over subintervals.

Transcript

00:01 So without showing you the graph, because you do have that given to you, i do want to point out that there's no reason to approximate the midpoint because you can get the exact value.

00:21 I like how, you know, in part a, at least they talk about, oh, you're going from one to seven with three sub -intervals.

00:31 So you can figure out the width of each, even though they tell you anyway, that you subtract your bounds divided by three.

00:39 So each width is two.

00:42 So your problem's going to be two times whatever the height is that we'll find in a second, two times the next, and then plus two times your third.

00:54 So here's your three rectangles, the width times the height, width times the height, width times the height.

00:59 Now i've had students in the past, so just guess whatever this number is.

01:04 But there's no need to guess because if we're starting at 1, okay, so the width is 2.

01:12 So 1 to 3, since we're doing midpoints, halfway between 1 and 3 is an x value of 2, or t value of 2, i guess i should be saying each time change in t instead of x.

01:27 But anyway, if you plug 2 in for the equation, we're looking at the square root of 20 because 10 times 2 is 20.

01:35 So instead of guessing like 4 .5, you can get an exact answer there.

01:41 Same thing with the next one.

01:42 So remember, the first one went from, first rectangle went from 1 to 3.

01:46 So the next one goes from 3 to 5, and that midpoint would be 4.

01:52 Well, now you can actually figure out, you know, we're at the square root of 40.

01:58 And then there's one more.

02:01 And be t equals six because it's between five and seven.

02:05 Well, if you plug it in there, we're looking at the square root of 60.

02:10 And some teachers might want, and they leave it like that.

02:15 You can actually simplify each of these because you can rewrite this one as four and five, two and two.

02:26 So two root five, so you can rewrite as four root five.

02:30 And then same thing with this one you know four and ten two and two two root ten so that would be four root ten and this one might even break down even more like six in ten what do we got three two two and five pair of twos so two root fifteen to bring a two in front so four root fifteen um anyway you can use a calculator to figure that you know approximation out...

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