Mixing
Minimum velocity
$begingroup$
There is a particle moving along the $x$-axis at any time $t geq 0$. The velocity of this particle is given by
$$v(t) = cos(πt) - t(6-2pi).$$
I am supposed to find the particle's minimum velocity over the interval $[0,2]$. I found the derivative of this velocity function and set it equal to zero, but I am having trouble with getting the right answer. I used the option on my calculator to find the min, but I can’t seen to find it using calculus. Help is greatly appreciated!
calculus derivatives
edited Jan 16 at 4:16
AEngineer
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
$endgroup$
add a comment |
$begingroup$
There is a particle moving along the $x$-axis at any time $t geq 0$. The velocity of this particle is given by
$$v(t) = cos(πt) - t(6-2pi).$$
I am supposed to find the particle's minimum velocity over the interval $[0,2]$. I found the derivative of this velocity function and set it equal to zero, but I am having trouble with getting the right answer. I used the option on my calculator to find the min, but I can’t seen to find it using calculus. Help is greatly appreciated!
calculus derivatives
edited Jan 16 at 4:16
AEngineer
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
$endgroup$
1
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34
add a comment |
$begingroup$
There is a particle moving along the $x$-axis at any time $t geq 0$. The velocity of this particle is given by
$$v(t) = cos(πt) - t(6-2pi).$$
I am supposed to find the particle's minimum velocity over the interval $[0,2]$. I found the derivative of this velocity function and set it equal to zero, but I am having trouble with getting the right answer. I used the option on my calculator to find the min, but I can’t seen to find it using calculus. Help is greatly appreciated!
calculus derivatives
edited Jan 16 at 4:16
AEngineer
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
$endgroup$
There is a particle moving along the $x$-axis at any time $t geq 0$. The velocity of this particle is given by
$$v(t) = cos(πt) - t(6-2pi).$$
I am supposed to find the particle's minimum velocity over the interval $[0,2]$. I found the derivative of this velocity function and set it equal to zero, but I am having trouble with getting the right answer. I used the option on my calculator to find the min, but I can’t seen to find it using calculus. Help is greatly appreciated!
calculus derivatives
calculus derivatives
edited Jan 16 at 4:16
AEngineer
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
edited Jan 16 at 4:16
AEngineer
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
edited Jan 16 at 4:16
AEngineer
1,5361217
edited Jan 16 at 4:16
AEngineer
1,5361217
edited Jan 16 at 4:16
AEngineer
1,5361217
1,5361217
asked Jan 16 at 3:47
math lovermath lover
53
asked Jan 16 at 3:47
math lovermath lover
53
asked Jan 16 at 3:47
math lovermath lover
53
53
1
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34
add a comment |
1
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34
1
1
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $tau$ when $v'(tau) = 0$ so that the minimum speed is given by $v(tau).$
$$v'(t) = -pi sin pi t + 2pi - 6 = 0.$$
So we have
begin{align}
sin pi t &= frac{2pi - 6}{pi} \
pi t &=
begin{cases}
arcsin (2 - 6/pi) + 2pi n \
pi - arcsin (2 - 6/pi) + 2pi n
end{cases}
end{align}
where $n in mathbb{Z}$ such that $t in [0,2]$. Taking the inverse sine results in 2 solutions over $2pi$ since there are 2 values of $y in (0,pi)$ that satisfy $sin y in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for
$$
tau = max
begin{cases}
frac{1}{pi} arcsinfrac{2pi - 6}{pi} \
1 -frac{1}{pi} arcsinfrac{2pi - 6}{pi}
end{cases}.
$$
Hope that helps.
Edit-
$y = arcsin a$ gives only 1 real $y$ for a given $a$, $|a| leq 1$ since it is, after all, a function; however, $sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.
$hskip 1 in$ 
If we say $sin(alpha) = b,$ then it's also true that $sin(alpha + beta) = b.$ But $2alpha + beta = pi,$ so $ sin(pi - alpha) = b.$ That is, $arcsin$ is restricted from $[-pi/2,pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2pi)$ that satisfy $sin y = b.$ Those angles are $arcsin b$ and $pi - arcsin b.$
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
$endgroup$
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
add a comment |
$begingroup$
The derivative would be $$v'(t) = -pi sin(pi t)-6+2pi$$
See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
$endgroup$
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
add a comment |
$begingroup$
The derivative of $v(t)$ is $$v'(t) = picosleft(pi tright) + 2pi - 6.$$
Setting the derivative equal to $0$ and solving for $t$, we get $t approx 0.31831$ which is within the interval $left[0, 2right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get
begin{align}
&v(0) = 1,\
&v(0.31831) = 0.630,\
&v(2) = 1.566.
end{align}
Thus, the minimum velocity over the interval $left[0, 2right]$ is $0.630$.
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075289%2fminimum-velocity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $tau$ when $v'(tau) = 0$ so that the minimum speed is given by $v(tau).$
$$v'(t) = -pi sin pi t + 2pi - 6 = 0.$$
So we have
begin{align}
sin pi t &= frac{2pi - 6}{pi} \
pi t &=
begin{cases}
arcsin (2 - 6/pi) + 2pi n \
pi - arcsin (2 - 6/pi) + 2pi n
end{cases}
end{align}
where $n in mathbb{Z}$ such that $t in [0,2]$. Taking the inverse sine results in 2 solutions over $2pi$ since there are 2 values of $y in (0,pi)$ that satisfy $sin y in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for
$$
tau = max
begin{cases}
frac{1}{pi} arcsinfrac{2pi - 6}{pi} \
1 -frac{1}{pi} arcsinfrac{2pi - 6}{pi}
end{cases}.
$$
Hope that helps.
Edit-
$y = arcsin a$ gives only 1 real $y$ for a given $a$, $|a| leq 1$ since it is, after all, a function; however, $sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.
$hskip 1 in$
If we say $sin(alpha) = b,$ then it's also true that $sin(alpha + beta) = b.$ But $2alpha + beta = pi,$ so $ sin(pi - alpha) = b.$ That is, $arcsin$ is restricted from $[-pi/2,pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2pi)$ that satisfy $sin y = b.$ Those angles are $arcsin b$ and $pi - arcsin b.$
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
$endgroup$
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
add a comment |
$begingroup$
The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $tau$ when $v'(tau) = 0$ so that the minimum speed is given by $v(tau).$
$$v'(t) = -pi sin pi t + 2pi - 6 = 0.$$
So we have
begin{align}
sin pi t &= frac{2pi - 6}{pi} \
pi t &=
begin{cases}
arcsin (2 - 6/pi) + 2pi n \
pi - arcsin (2 - 6/pi) + 2pi n
end{cases}
end{align}
where $n in mathbb{Z}$ such that $t in [0,2]$. Taking the inverse sine results in 2 solutions over $2pi$ since there are 2 values of $y in (0,pi)$ that satisfy $sin y in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for
$$
tau = max
begin{cases}
frac{1}{pi} arcsinfrac{2pi - 6}{pi} \
1 -frac{1}{pi} arcsinfrac{2pi - 6}{pi}
end{cases}.
$$
Hope that helps.
Edit-
$y = arcsin a$ gives only 1 real $y$ for a given $a$, $|a| leq 1$ since it is, after all, a function; however, $sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.
$hskip 1 in$
If we say $sin(alpha) = b,$ then it's also true that $sin(alpha + beta) = b.$ But $2alpha + beta = pi,$ so $ sin(pi - alpha) = b.$ That is, $arcsin$ is restricted from $[-pi/2,pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2pi)$ that satisfy $sin y = b.$ Those angles are $arcsin b$ and $pi - arcsin b.$
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
$endgroup$
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
add a comment |
$begingroup$
The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $tau$ when $v'(tau) = 0$ so that the minimum speed is given by $v(tau).$
$$v'(t) = -pi sin pi t + 2pi - 6 = 0.$$
So we have
begin{align}
sin pi t &= frac{2pi - 6}{pi} \
pi t &=
begin{cases}
arcsin (2 - 6/pi) + 2pi n \
pi - arcsin (2 - 6/pi) + 2pi n
end{cases}
end{align}
where $n in mathbb{Z}$ such that $t in [0,2]$. Taking the inverse sine results in 2 solutions over $2pi$ since there are 2 values of $y in (0,pi)$ that satisfy $sin y in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for
$$
tau = max
begin{cases}
frac{1}{pi} arcsinfrac{2pi - 6}{pi} \
1 -frac{1}{pi} arcsinfrac{2pi - 6}{pi}
end{cases}.
$$
Hope that helps.
Edit-
$y = arcsin a$ gives only 1 real $y$ for a given $a$, $|a| leq 1$ since it is, after all, a function; however, $sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.
$hskip 1 in$
If we say $sin(alpha) = b,$ then it's also true that $sin(alpha + beta) = b.$ But $2alpha + beta = pi,$ so $ sin(pi - alpha) = b.$ That is, $arcsin$ is restricted from $[-pi/2,pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2pi)$ that satisfy $sin y = b.$ Those angles are $arcsin b$ and $pi - arcsin b.$
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
$endgroup$
The lowest velocity occurs right as acceleration changes direction. That is, we need to find the time $tau$ when $v'(tau) = 0$ so that the minimum speed is given by $v(tau).$
$$v'(t) = -pi sin pi t + 2pi - 6 = 0.$$
So we have
begin{align}
sin pi t &= frac{2pi - 6}{pi} \
pi t &=
begin{cases}
arcsin (2 - 6/pi) + 2pi n \
pi - arcsin (2 - 6/pi) + 2pi n
end{cases}
end{align}
where $n in mathbb{Z}$ such that $t in [0,2]$. Taking the inverse sine results in 2 solutions over $2pi$ since there are 2 values of $y in (0,pi)$ that satisfy $sin y in [0,1).$ By inspection, we can see that $n = 0,$ otherwise $t notin [0,2].$ Moving with this, we can say that due to the $-kt$ term in $v(t)$, we are looking for
$$
tau = max
begin{cases}
frac{1}{pi} arcsinfrac{2pi - 6}{pi} \
1 -frac{1}{pi} arcsinfrac{2pi - 6}{pi}
end{cases}.
$$
Hope that helps.
Edit-
$y = arcsin a$ gives only 1 real $y$ for a given $a$, $|a| leq 1$ since it is, after all, a function; however, $sin y = a$ has infinitely many real $y$ for such an $a$. It's better illustrated with a picture.
$hskip 1 in$
If we say $sin(alpha) = b,$ then it's also true that $sin(alpha + beta) = b.$ But $2alpha + beta = pi,$ so $ sin(pi - alpha) = b.$ That is, $arcsin$ is restricted from $[-pi/2,pi/2]$ to guarantee that it is a function, but there are 2 angles $y$ in $[0,2pi)$ that satisfy $sin y = b.$ Those angles are $arcsin b$ and $pi - arcsin b.$
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
edited Jan 16 at 6:22
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
answered Jan 16 at 4:43
AEngineerAEngineer
1,5361217
1,5361217
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
add a comment |
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
What did you use when you split the function into two functions?
$endgroup$
– math lover
Jan 16 at 5:31
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
$begingroup$
Does that make a little more sense? Let me know.
$endgroup$
– AEngineer
Jan 16 at 6:23
1
1
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
$begingroup$
Yes! Thank you!
$endgroup$
– math lover
Jan 16 at 6:39
add a comment |
$begingroup$
The derivative would be $$v'(t) = -pi sin(pi t)-6+2pi$$
See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
$endgroup$
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
add a comment |
$begingroup$
The derivative would be $$v'(t) = -pi sin(pi t)-6+2pi$$
See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
$endgroup$
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
add a comment |
$begingroup$
The derivative would be $$v'(t) = -pi sin(pi t)-6+2pi$$
See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
$endgroup$
The derivative would be $$v'(t) = -pi sin(pi t)-6+2pi$$
See if this is the derivative you were coming up with, then set $v'(t) = 0$ and solve for $t$. Check to see if these critical values are local minimum(s) or local maximum(s). Also remember to check your endpoints by plugging them back into $v(t)$
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
answered Jan 16 at 4:44
WaveXWaveX
2,6922722
2,6922722
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
add a comment |
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
I did get that derivative and I followed those steps. I am getting the answer .0287 for t, but I am not sure if this is correct because my graphing calculator said that the min was at t= .9713. Can you verify?
$endgroup$
– math lover
Jan 16 at 5:24
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
You mention you're using a calculator...are you sure you are in the correct mode since you're using $sin$? You're in radians and not degrees?
$endgroup$
– WaveX
Jan 16 at 5:28
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Yes, I am in radians.
$endgroup$
– math lover
Jan 16 at 5:30
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Both of those values equal $0$ when plugged into the derivative. Make sure you are choosing the one that gives the absolute minimum in the given range
$endgroup$
– WaveX
Jan 16 at 5:35
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
$begingroup$
Is interval the same as range?
$endgroup$
– math lover
Jan 16 at 5:41
add a comment |
$begingroup$
The derivative of $v(t)$ is $$v'(t) = picosleft(pi tright) + 2pi - 6.$$
Setting the derivative equal to $0$ and solving for $t$, we get $t approx 0.31831$ which is within the interval $left[0, 2right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get
begin{align}
&v(0) = 1,\
&v(0.31831) = 0.630,\
&v(2) = 1.566.
end{align}
Thus, the minimum velocity over the interval $left[0, 2right]$ is $0.630$.
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
add a comment |
$begingroup$
The derivative of $v(t)$ is $$v'(t) = picosleft(pi tright) + 2pi - 6.$$
Setting the derivative equal to $0$ and solving for $t$, we get $t approx 0.31831$ which is within the interval $left[0, 2right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get
begin{align}
&v(0) = 1,\
&v(0.31831) = 0.630,\
&v(2) = 1.566.
end{align}
Thus, the minimum velocity over the interval $left[0, 2right]$ is $0.630$.
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
add a comment |
$begingroup$
The derivative of $v(t)$ is $$v'(t) = picosleft(pi tright) + 2pi - 6.$$
Setting the derivative equal to $0$ and solving for $t$, we get $t approx 0.31831$ which is within the interval $left[0, 2right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get
begin{align}
&v(0) = 1,\
&v(0.31831) = 0.630,\
&v(2) = 1.566.
end{align}
Thus, the minimum velocity over the interval $left[0, 2right]$ is $0.630$.
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
$endgroup$
The derivative of $v(t)$ is $$v'(t) = picosleft(pi tright) + 2pi - 6.$$
Setting the derivative equal to $0$ and solving for $t$, we get $t approx 0.31831$ which is within the interval $left[0, 2right]$. However, as John Douma pointed out, you also need to check the endpoints, $t = 0$ and $t = 2$. Simply evaluating, we get
begin{align}
&v(0) = 1,\
&v(0.31831) = 0.630,\
&v(2) = 1.566.
end{align}
Thus, the minimum velocity over the interval $left[0, 2right]$ is $0.630$.
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
answered Jan 16 at 4:48
PiKindOfGuyPiKindOfGuy
18611
18611
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075289%2fminimum-velocity%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
1
$begingroup$
Remember that the derivative will only give you local extrema. You must also consider the boundaries.
$endgroup$
– John Douma
Jan 16 at 4:34