the police department has a machine that makes a graph for each car that drives on a certain road, graphing…

the police department has a machine that makes a graph for each car that drives on a certain road, graphing the position of the car (in miles) as a function of time (in minutes).\n1. suppose the six graphs from the week 1 group project are graphs that this machine recorded. for cars (a), (b), and (c) from week 1 calculate the average velocity on the interval 2,3 i.e. between 2 and 3 minutes. then calculate the average velocity over 2,2.5. which calculation seems closer to the instantaneous velocity at 2 minutes?\n2. one way of estimating the instantaneous velocity at 2 minutes is by calculating average velocities over 2,2 + h, and choosing smaller and smaller values for the interval length h. you started this process already in question #1. (h was 1 for the interval 2,3 and.5 for the interval 2,2.5.) continue to carry out this strategy for one more smaller interval on cars (a), (b), and (c). use the results from questions #1 and #2 to approximate what the instantaneous velocity at 2 minutes is for each car.\n3. create a graph of a position function where the average velocity over 2,3 is a better estimate for the instantaneous velocity at 2 minutes than the average velocity over 2,2.5.\n4. determine whether the following statements are true and give an explanation or counterexample.\na. the value of lim(x→3) (x² - 9)/(x - 3) does not exist.\nb. the value of lim(x→a) f(x) can always be found by computing f(a).\nc. the value of lim(x→a) f(x) does not exist if f(a) is undefined.

the police department has a machine that makes a graph for each car that drives on a certain road, graphing the position of the car (in miles) as a function of time (in minutes).\n1. suppose the six graphs from the week 1 group project are graphs that this machine recorded. for cars (a), (b), and (c) from week 1 calculate the average velocity on the interval 2,3 i.e. between 2 and 3 minutes. then calculate the average velocity over 2,2.5. which calculation seems closer to the instantaneous velocity at 2 minutes?\n2. one way of estimating the instantaneous velocity at 2 minutes is by calculating average velocities over 2,2 + h, and choosing smaller and smaller values for the interval length h. you started this process already in question #1. (h was 1 for the interval 2,3 and.5 for the interval 2,2.5.) continue to carry out this strategy for one more smaller interval on cars (a), (b), and (c). use the results from questions #1 and #2 to approximate what the instantaneous velocity at 2 minutes is for each car.\n3. create a graph of a position function where the average velocity over 2,3 is a better estimate for the instantaneous velocity at 2 minutes than the average velocity over 2,2.5.\n4. determine whether the following statements are true and give an explanation or counterexample.\na. the value of lim(x→3) (x² - 9)/(x - 3) does not exist.\nb. the value of lim(x→a) f(x) can always be found by computing f(a).\nc. the value of lim(x→a) f(x) does not exist if f(a) is undefined.

Answer

Explanation:

Step1: Recall average - velocity formula

The average velocity $v_{avg}$ of an object over the interval $[a,b]$ for a position - function $s(t)$ is given by $v_{avg}=\frac{s(b)-s(a)}{b - a}$.

Step2: Calculate average velocity over $[2,3]$ for cars (A), (B), and (C)

Let $s_A(t)$, $s_B(t)$, and $s_C(t)$ be the position - functions of cars (A), (B), and (C) respectively. Then $v_{A,[2,3]}=\frac{s_A(3)-s_A(2)}{3 - 2}=s_A(3)-s_A(2)$, $v_{B,[2,3]}=s_B(3)-s_B(2)$, $v_{C,[2,3]}=s_C(3)-s_C(2)$.

Step3: Calculate average velocity over $[2,2.5]$ for cars (A), (B), and (C)

$v_{A,[2,2.5]}=\frac{s_A(2.5)-s_A(2)}{2.5 - 2}=\frac{s_A(2.5)-s_A(2)}{0.5}=2(s_A(2.5)-s_A(2))$, $v_{B,[2,2.5]}=2(s_B(2.5)-s_B(2))$, $v_{C,[2,2.5]}=2(s_C(2.5)-s_C(2))$. The average velocity over the smaller interval $[2,2.5]$ is closer to the instantaneous velocity at $t = 2$ because the smaller the interval, the better the average velocity approximates the instantaneous velocity.

Step4: Choose a smaller interval (e.g., $[2,2.1]$)

For car (A): $v_{A,[2,2.1]}=\frac{s_A(2.1)-s_A(2)}{2.1 - 2}=10(s_A(2.1)-s_A(2))$. Similarly for cars (B) and (C). As we take smaller and smaller intervals around $t = 2$, the average velocities will approach the instantaneous velocity at $t = 2$.

Step5: Analyze limit statements

a.

We simplify $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}=\lim_{x\rightarrow3}\frac{(x + 3)(x - 3)}{x - 3}=\lim_{x\rightarrow3}(x + 3)=6$. So the statement "The value of $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}$ does not exist" is false.

b.

The value of $\lim_{x\rightarrow a}f(x)$ cannot always be found by computing $f(a)$. For example, if $f(x)=\frac{x^{2}-9}{x - 3}$ and $a = 3$, $f(3)$ is undefined but $\lim_{x\rightarrow3}f(x)=6$. So this statement is false.

c.

The value of $\lim_{x\rightarrow a}f(x)$ can exist even if $f(a)$ is undefined. As in the previous example of $f(x)=\frac{x^{2}-9}{x - 3}$ at $a = 3$, $\lim_{x\rightarrow3}f(x)=6$ while $f(3)$ is undefined. So this statement is false.

Answer:

  1. Average velocities for cars (A), (B), (C) over $[2,3]$ and $[2,2.5]$ are calculated as above. The average velocity over $[2,2.5]$ is closer to the instantaneous velocity at $t = 2$.
  2. By choosing a smaller interval like $[2,2.1]$ and calculating average velocities, we can approximate the instantaneous velocity at $t = 2$ for each car.
  3. A graph of a position - function with a relatively constant slope near $t = 2$ in the interval $[2,3]$ compared to $[2,2.5]$ can be created. For example, a linear function $s(t)=mt + b$ with a constant slope $m$ in the relevant interval.

a. False. $\lim_{x\rightarrow3}\frac{x^{2}-9}{x - 3}=6$. b. False. There are cases where $f(a)$ is not defined but $\lim_{x\rightarrow a}f(x)$ exists. c. False. $\lim_{x\rightarrow a}f(x)$ can exist when $f(a)$ is undefined.