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Using the Mean Value Theorem to show that there exists some point on a graph whose tangent has a given slope.
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I have been reading Calculus with Analytical Geometry by George F. Simmons. One of the practice problems is as follows:
A car starts from rest and travels 4mi along a straight road in 6 minutes. Use the Mean Value Theorem to show that at some moment during the trip its speed was exactly 40 mi/h.
Generally, to find a point $c$ that satisfies the MVT for some function $f(x)$, I solve the following:
$$f'(c) = fracf(b) - f(a)b-a$$
However, I'm unsure what function $f(x)$ to construct that will fit the terms of the problem other than $f(x) = frac23x$, which is equivalent to a vehicle traveling at $frac23$ a mile per minute.
The derivative of that function, however, is a constant, which is not useful.
Thank you.
calculus derivatives
asked Aug 1 at 19:12
Jamie Corkhill
1158
add a comment |Â
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0
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I have been reading Calculus with Analytical Geometry by George F. Simmons. One of the practice problems is as follows:
A car starts from rest and travels 4mi along a straight road in 6 minutes. Use the Mean Value Theorem to show that at some moment during the trip its speed was exactly 40 mi/h.
Generally, to find a point $c$ that satisfies the MVT for some function $f(x)$, I solve the following:
$$f'(c) = fracf(b) - f(a)b-a$$
However, I'm unsure what function $f(x)$ to construct that will fit the terms of the problem other than $f(x) = frac23x$, which is equivalent to a vehicle traveling at $frac23$ a mile per minute.
The derivative of that function, however, is a constant, which is not useful.
Thank you.
calculus derivatives
asked Aug 1 at 19:12
Jamie Corkhill
1158
Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I have been reading Calculus with Analytical Geometry by George F. Simmons. One of the practice problems is as follows:
A car starts from rest and travels 4mi along a straight road in 6 minutes. Use the Mean Value Theorem to show that at some moment during the trip its speed was exactly 40 mi/h.
Generally, to find a point $c$ that satisfies the MVT for some function $f(x)$, I solve the following:
$$f'(c) = fracf(b) - f(a)b-a$$
However, I'm unsure what function $f(x)$ to construct that will fit the terms of the problem other than $f(x) = frac23x$, which is equivalent to a vehicle traveling at $frac23$ a mile per minute.
The derivative of that function, however, is a constant, which is not useful.
Thank you.
calculus derivatives
asked Aug 1 at 19:12
Jamie Corkhill
1158
I have been reading Calculus with Analytical Geometry by George F. Simmons. One of the practice problems is as follows:
A car starts from rest and travels 4mi along a straight road in 6 minutes. Use the Mean Value Theorem to show that at some moment during the trip its speed was exactly 40 mi/h.
Generally, to find a point $c$ that satisfies the MVT for some function $f(x)$, I solve the following:
$$f'(c) = fracf(b) - f(a)b-a$$
However, I'm unsure what function $f(x)$ to construct that will fit the terms of the problem other than $f(x) = frac23x$, which is equivalent to a vehicle traveling at $frac23$ a mile per minute.
The derivative of that function, however, is a constant, which is not useful.
Thank you.
calculus derivatives
asked Aug 1 at 19:12
Jamie Corkhill
1158
asked Aug 1 at 19:12
Jamie Corkhill
1158
asked Aug 1 at 19:12
Jamie Corkhill
1158
asked Aug 1 at 19:12
Jamie Corkhill
1158
1158
Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23
add a comment |Â
Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23
Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
2
down vote
accepted
MVT state that for an interval $[a,b]$ there is c such that $f^prime(c)=dfracf(b)-f(a)b-a$
Let us calculate the car's speed in mph
$$dfrac4mbox miles6mbox minutes=dfracxmbox miles60mbox minutes$$
$$x=40mbox mph$$
This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit:
Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$dfracf(6)-f(0)6-0=dfrac23dfracmbox milesmbox minute$ which is $40$ mph
MVT states that at some $0le cle6,f^prime(x)=dfrac23$
answered Aug 1 at 19:22
Key Flex
3,787422
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
add a comment |Â
up vote
0
down vote
You have $a=0 texth$ and $b=6textmin=0.1texth$ and
$f(a) = 0 textmi $ and $f(b) = 4textmi$, this gives
$$
f'(c) = fracf(b)-f(a)b-a
= frac4 - 00.1 - 0 fractextmitexth= 40 fractextmitexth
$$
for some $c in (a,b) = (0 texth, 0.1 texth)$.
As $f$ is a distance and its variable is a time, $f'$ is a velocity.
answered Aug 1 at 19:26
mvw
30.2k22250
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
MVT state that for an interval $[a,b]$ there is c such that $f^prime(c)=dfracf(b)-f(a)b-a$
Let us calculate the car's speed in mph
$$dfrac4mbox miles6mbox minutes=dfracxmbox miles60mbox minutes$$
$$x=40mbox mph$$
This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit:
Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$dfracf(6)-f(0)6-0=dfrac23dfracmbox milesmbox minute$ which is $40$ mph
MVT states that at some $0le cle6,f^prime(x)=dfrac23$
answered Aug 1 at 19:22
Key Flex
3,787422
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
add a comment |Â
up vote
2
down vote
accepted
MVT state that for an interval $[a,b]$ there is c such that $f^prime(c)=dfracf(b)-f(a)b-a$
Let us calculate the car's speed in mph
$$dfrac4mbox miles6mbox minutes=dfracxmbox miles60mbox minutes$$
$$x=40mbox mph$$
This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit:
Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$dfracf(6)-f(0)6-0=dfrac23dfracmbox milesmbox minute$ which is $40$ mph
MVT states that at some $0le cle6,f^prime(x)=dfrac23$
answered Aug 1 at 19:22
Key Flex
3,787422
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
MVT state that for an interval $[a,b]$ there is c such that $f^prime(c)=dfracf(b)-f(a)b-a$
Let us calculate the car's speed in mph
$$dfrac4mbox miles6mbox minutes=dfracxmbox miles60mbox minutes$$
$$x=40mbox mph$$
This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit:
Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$dfracf(6)-f(0)6-0=dfrac23dfracmbox milesmbox minute$ which is $40$ mph
MVT states that at some $0le cle6,f^prime(x)=dfrac23$
answered Aug 1 at 19:22
Key Flex
3,787422
MVT state that for an interval $[a,b]$ there is c such that $f^prime(c)=dfracf(b)-f(a)b-a$
Let us calculate the car's speed in mph
$$dfrac4mbox miles6mbox minutes=dfracxmbox miles60mbox minutes$$
$$x=40mbox mph$$
This is the Average speed.
We need to show that at some time, the car's Average speed is equaled to the Instantaneous speed. By MVT we can say that average rate of change is, at some time, is equal to the instantaneous rate of change. Hence proved.
Edit:
Letting $f(t)$ be distance at time $t$ minutes,
$f(0)=0$
$f(6)=4$
$dfracf(6)-f(0)6-0=dfrac23dfracmbox milesmbox minute$ which is $40$ mph
MVT states that at some $0le cle6,f^prime(x)=dfrac23$
answered Aug 1 at 19:22
Key Flex
3,787422
edited Aug 1 at 19:33
answered Aug 1 at 19:22
Key Flex
3,787422
answered Aug 1 at 19:22
Key Flex
3,787422
answered Aug 1 at 19:22
Key Flex
3,787422
3,787422
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
add a comment |Â
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
Yes, you just let "f(t)" be the position of the car at time t during the trip, let "a" be the time at the start of the trip, and let "b" be the time at the end of the trip. Then f'(t) is the velocity of the car at time t during the trip. f(b) - f(a) is the distance traveled (4 miles), b - a is the time taken (6 minutes) and the rest of the proof continues as @Key Flex shows in this answer. Basically, MVT tells you that there is some c between a and b (during the trip, not before it or after it) where the car was (instantaneously) going the average speed. And the average speed is 40 mph.
– Jake Griffin
Aug 1 at 19:26
add a comment |Â
up vote
0
down vote
You have $a=0 texth$ and $b=6textmin=0.1texth$ and
$f(a) = 0 textmi $ and $f(b) = 4textmi$, this gives
$$
f'(c) = fracf(b)-f(a)b-a
= frac4 - 00.1 - 0 fractextmitexth= 40 fractextmitexth
$$
for some $c in (a,b) = (0 texth, 0.1 texth)$.
As $f$ is a distance and its variable is a time, $f'$ is a velocity.
answered Aug 1 at 19:26
mvw
30.2k22250
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
add a comment |Â
up vote
0
down vote
You have $a=0 texth$ and $b=6textmin=0.1texth$ and
$f(a) = 0 textmi $ and $f(b) = 4textmi$, this gives
$$
f'(c) = fracf(b)-f(a)b-a
= frac4 - 00.1 - 0 fractextmitexth= 40 fractextmitexth
$$
for some $c in (a,b) = (0 texth, 0.1 texth)$.
As $f$ is a distance and its variable is a time, $f'$ is a velocity.
answered Aug 1 at 19:26
mvw
30.2k22250
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
add a comment |Â
up vote
0
down vote
up vote
0
down vote
You have $a=0 texth$ and $b=6textmin=0.1texth$ and
$f(a) = 0 textmi $ and $f(b) = 4textmi$, this gives
$$
f'(c) = fracf(b)-f(a)b-a
= frac4 - 00.1 - 0 fractextmitexth= 40 fractextmitexth
$$
for some $c in (a,b) = (0 texth, 0.1 texth)$.
As $f$ is a distance and its variable is a time, $f'$ is a velocity.
answered Aug 1 at 19:26
mvw
30.2k22250
You have $a=0 texth$ and $b=6textmin=0.1texth$ and
$f(a) = 0 textmi $ and $f(b) = 4textmi$, this gives
$$
f'(c) = fracf(b)-f(a)b-a
= frac4 - 00.1 - 0 fractextmitexth= 40 fractextmitexth
$$
for some $c in (a,b) = (0 texth, 0.1 texth)$.
As $f$ is a distance and its variable is a time, $f'$ is a velocity.
answered Aug 1 at 19:26
mvw
30.2k22250
answered Aug 1 at 19:26
mvw
30.2k22250
answered Aug 1 at 19:26
mvw
30.2k22250
answered Aug 1 at 19:26
mvw
30.2k22250
30.2k22250
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
add a comment |Â
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
Does "mvw" stand for "Mean Value Wizard"? ;-)
– Jake Griffin
Aug 1 at 19:31
nah its just initials
– mvw
Aug 1 at 19:32
nah its just initials
– mvw
Aug 1 at 19:32
add a comment |Â
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Hint: Let $s(t)$ be the distance in miles traveled over number of hours $t$. Then the derivative of the function, $s'(t)$ gives you its speed at any given time during the trip.
– packetpacket
Aug 1 at 19:17
You don't need to construct the function. The mean value theorem says that if the function is smooth, and you know the endpoints, then somewhere in the interval the derivative equals the average rate of change. You do not need to solve for $c$ just acknowledge that it exists. Also, I suggest you convert the time units to hours.
– Doug M
Aug 1 at 19:23