What is the Range of $5|sin x|+12|cos x|$
$begingroup$
What is the Range of $5|sin x|+12|cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
functions
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
$endgroup$
add a comment |
$begingroup$
What is the Range of $5|sin x|+12|cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
functions
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
$endgroup$
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42
add a comment |
$begingroup$
What is the Range of $5|sin x|+12|cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
functions
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
$endgroup$
What is the Range of $5|sin x|+12|cos x|$ ?
I entered the value in desmos.com and getting the range as $[5,13]$.
Using $sqrt{5^2+12^2} =13$, i am able to get maximum value but not able to find the minimum.
functions
functions
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
edited Dec 15 '18 at 4:33
Kemono Chen
3,0771743
3,0771743
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
asked Dec 15 '18 at 2:39
Samar Imam ZaidiSamar Imam Zaidi
1,5501519
1,5501519
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42
add a comment |
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Without loss of generality we may assume $0le thetalepi/2$ so $cos theta$ and $sin theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0lethetalefrac{pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(frac{5}{12}x)^2=1$ to get $x=frac{12}{13}$ and $y=frac{5}{13}$, which yields a maximum of $f(frac{12}{13},frac{5}{13})=13$.
Therefore the range is $[5,13]$.
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
$endgroup$
add a comment |
$begingroup$
If $f(x) = 5|sin x| + 12 |cos x|$, then
begin{align*}
f(x) &= sqrt{f(x)^2} \
&= sqrt{25 sin^2 x + 144 cos^2 x + 60 |sin x cos x|} \
&= sqrt{25 + (144 - 25) cos^2 x + 60 |sin x cos x|} \
&ge 5
end{align*}
with equality obtained when $cos x = 0$.
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
$endgroup$
add a comment |
$begingroup$
The four quadrants give the four combinations of signs of $sin$ and $cos$. Let us work initially in the first quadrant, where both functions are positive. We can then remove the absolute value signs, take a derivative, and set to zero.
$$frac d{dx}(5 sin x + 12 cos x)=5cos x - 12 sin x$$
This is zero when $tan x=frac 5{12}$, giving the maximum you found. The minimum must then come at one end of the interval, and if you check $x=0, frac pi 2$ you find the minimum at $frac pi 2$, which is $5$. You can do the same in the other four quadrants, flipping the signs of $sin x$ and $cos x$ as required, and find that the minimum is $5$ again.
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
Another possible approach.
For the first quadrant: $5sin(x) + 12cos(x) = 13sin(x + arccos(frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
$endgroup$
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%2f3040110%2fwhat-is-the-range-of-5-sin-x12-cos-x%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Without loss of generality we may assume $0le thetalepi/2$ so $cos theta$ and $sin theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0lethetalefrac{pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(frac{5}{12}x)^2=1$ to get $x=frac{12}{13}$ and $y=frac{5}{13}$, which yields a maximum of $f(frac{12}{13},frac{5}{13})=13$.
Therefore the range is $[5,13]$.
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
$endgroup$
add a comment |
$begingroup$
Without loss of generality we may assume $0le thetalepi/2$ so $cos theta$ and $sin theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0lethetalefrac{pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(frac{5}{12}x)^2=1$ to get $x=frac{12}{13}$ and $y=frac{5}{13}$, which yields a maximum of $f(frac{12}{13},frac{5}{13})=13$.
Therefore the range is $[5,13]$.
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
$endgroup$
add a comment |
$begingroup$
Without loss of generality we may assume $0le thetalepi/2$ so $cos theta$ and $sin theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0lethetalefrac{pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(frac{5}{12}x)^2=1$ to get $x=frac{12}{13}$ and $y=frac{5}{13}$, which yields a maximum of $f(frac{12}{13},frac{5}{13})=13$.
Therefore the range is $[5,13]$.
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
$endgroup$
Without loss of generality we may assume $0le thetalepi/2$ so $cos theta$ and $sin theta$ are positive.
You can solve this with calculus (set the derivative equal to zero and solve for $theta$), but here is a geometric way to think about the problem which avoids calculus.
The level curves of the scalar function $f(x,y)=5y+12x$ are the parallel lines
$$ 5y+12x=C $$
for various real values $C$. One may "parametrize" such lines by using a perpendicular line through the origin, namely the line $y=frac{5}{12}x$. Since the slope is less than $1$, the perpendicular line is tilted towards the $x$-axis, so the "first" of level curves to intersect the unit circle arc $0lethetalefrac{pi}{2}$ does so at the point $(0,1)$, which corresponds to an output of $f(0,1)=5$. The "last" of the level curves to intersect the arc occurs where the perpendicular line intersects it, so solve $x^2+(frac{5}{12}x)^2=1$ to get $x=frac{12}{13}$ and $y=frac{5}{13}$, which yields a maximum of $f(frac{12}{13},frac{5}{13})=13$.
Therefore the range is $[5,13]$.
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
answered Dec 15 '18 at 2:58
arctic ternarctic tern
11.9k31535
11.9k31535
add a comment |
add a comment |
$begingroup$
If $f(x) = 5|sin x| + 12 |cos x|$, then
begin{align*}
f(x) &= sqrt{f(x)^2} \
&= sqrt{25 sin^2 x + 144 cos^2 x + 60 |sin x cos x|} \
&= sqrt{25 + (144 - 25) cos^2 x + 60 |sin x cos x|} \
&ge 5
end{align*}
with equality obtained when $cos x = 0$.
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
$endgroup$
add a comment |
$begingroup$
If $f(x) = 5|sin x| + 12 |cos x|$, then
begin{align*}
f(x) &= sqrt{f(x)^2} \
&= sqrt{25 sin^2 x + 144 cos^2 x + 60 |sin x cos x|} \
&= sqrt{25 + (144 - 25) cos^2 x + 60 |sin x cos x|} \
&ge 5
end{align*}
with equality obtained when $cos x = 0$.
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
$endgroup$
add a comment |
$begingroup$
If $f(x) = 5|sin x| + 12 |cos x|$, then
begin{align*}
f(x) &= sqrt{f(x)^2} \
&= sqrt{25 sin^2 x + 144 cos^2 x + 60 |sin x cos x|} \
&= sqrt{25 + (144 - 25) cos^2 x + 60 |sin x cos x|} \
&ge 5
end{align*}
with equality obtained when $cos x = 0$.
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
$endgroup$
If $f(x) = 5|sin x| + 12 |cos x|$, then
begin{align*}
f(x) &= sqrt{f(x)^2} \
&= sqrt{25 sin^2 x + 144 cos^2 x + 60 |sin x cos x|} \
&= sqrt{25 + (144 - 25) cos^2 x + 60 |sin x cos x|} \
&ge 5
end{align*}
with equality obtained when $cos x = 0$.
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
edited Dec 15 '18 at 2:54
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
answered Dec 15 '18 at 2:49
T. BongersT. Bongers
23.3k54662
23.3k54662
add a comment |
add a comment |
$begingroup$
The four quadrants give the four combinations of signs of $sin$ and $cos$. Let us work initially in the first quadrant, where both functions are positive. We can then remove the absolute value signs, take a derivative, and set to zero.
$$frac d{dx}(5 sin x + 12 cos x)=5cos x - 12 sin x$$
This is zero when $tan x=frac 5{12}$, giving the maximum you found. The minimum must then come at one end of the interval, and if you check $x=0, frac pi 2$ you find the minimum at $frac pi 2$, which is $5$. You can do the same in the other four quadrants, flipping the signs of $sin x$ and $cos x$ as required, and find that the minimum is $5$ again.
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
The four quadrants give the four combinations of signs of $sin$ and $cos$. Let us work initially in the first quadrant, where both functions are positive. We can then remove the absolute value signs, take a derivative, and set to zero.
$$frac d{dx}(5 sin x + 12 cos x)=5cos x - 12 sin x$$
This is zero when $tan x=frac 5{12}$, giving the maximum you found. The minimum must then come at one end of the interval, and if you check $x=0, frac pi 2$ you find the minimum at $frac pi 2$, which is $5$. You can do the same in the other four quadrants, flipping the signs of $sin x$ and $cos x$ as required, and find that the minimum is $5$ again.
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
add a comment |
$begingroup$
The four quadrants give the four combinations of signs of $sin$ and $cos$. Let us work initially in the first quadrant, where both functions are positive. We can then remove the absolute value signs, take a derivative, and set to zero.
$$frac d{dx}(5 sin x + 12 cos x)=5cos x - 12 sin x$$
This is zero when $tan x=frac 5{12}$, giving the maximum you found. The minimum must then come at one end of the interval, and if you check $x=0, frac pi 2$ you find the minimum at $frac pi 2$, which is $5$. You can do the same in the other four quadrants, flipping the signs of $sin x$ and $cos x$ as required, and find that the minimum is $5$ again.
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
$endgroup$
The four quadrants give the four combinations of signs of $sin$ and $cos$. Let us work initially in the first quadrant, where both functions are positive. We can then remove the absolute value signs, take a derivative, and set to zero.
$$frac d{dx}(5 sin x + 12 cos x)=5cos x - 12 sin x$$
This is zero when $tan x=frac 5{12}$, giving the maximum you found. The minimum must then come at one end of the interval, and if you check $x=0, frac pi 2$ you find the minimum at $frac pi 2$, which is $5$. You can do the same in the other four quadrants, flipping the signs of $sin x$ and $cos x$ as required, and find that the minimum is $5$ again.
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
answered Dec 15 '18 at 2:53
Ross MillikanRoss Millikan
297k23198371
297k23198371
add a comment |
add a comment |
$begingroup$
Another possible approach.
For the first quadrant: $5sin(x) + 12cos(x) = 13sin(x + arccos(frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
$endgroup$
add a comment |
$begingroup$
Another possible approach.
For the first quadrant: $5sin(x) + 12cos(x) = 13sin(x + arccos(frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
$endgroup$
add a comment |
$begingroup$
Another possible approach.
For the first quadrant: $5sin(x) + 12cos(x) = 13sin(x + arccos(frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
$endgroup$
Another possible approach.
For the first quadrant: $5sin(x) + 12cos(x) = 13sin(x + arccos(frac{5}{13}))$. Can follow from there for the rest of the quadrants. Can also try alternative forms for arguments in order to adapt to the values of $x$.
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
answered Dec 15 '18 at 3:03
MakinaMakina
1,1651316
1,1651316
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%2f3040110%2fwhat-is-the-range-of-5-sin-x12-cos-x%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
$begingroup$
Manually i get the minimum value at $x=frac{π}{2}$ , but how do i get it by solving
$endgroup$
– Samar Imam Zaidi
Dec 15 '18 at 2:42