Mixing
Which point of the graph of $y=sqrt{x}$ is closest to the point $(1,0)$?
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This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=sqrt{(x-1)^2+(y-0)^2}$
$d=sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
calculus
edited Jan 17 at 6:13
Paras Khosla
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
$endgroup$
|
show 2 more comments
$begingroup$
This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=sqrt{(x-1)^2+(y-0)^2}$
$d=sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
calculus
edited Jan 17 at 6:13
Paras Khosla
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
$endgroup$
$begingroup$
math.stackexchange.com/questions/1531154/…
$endgroup$
– dimebucker
Jan 17 at 5:53
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
1
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
1
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
3
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37
|
show 2 more comments
$begingroup$
This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=sqrt{(x-1)^2+(y-0)^2}$
$d=sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
calculus
edited Jan 17 at 6:13
Paras Khosla
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
$endgroup$
This problem was assigned for an AP Calculus AB class and was not allowed a calculator:
Which point of the graph of $y=sqrt{x}$ is closest to the point $(1,0)$?
We are not given answers and the teacher will be absent for $2$ weeks. I need to check my answer $(1/2,1/4)$ before she returns.
Here is my work:
$d=sqrt{(x-1)^2+(y-0)^2}$
$d=sqrt{(x-1)^2+x}$
$d^2=(x-1)^2+x$
$2dd'=2(x-1)+1$
$d'=(2(x-1)+1)/(2d)$
$0=2(x-1)+1$
$x=1/2$
calculus
calculus
edited Jan 17 at 6:13
Paras Khosla
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
edited Jan 17 at 6:13
Paras Khosla
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
edited Jan 17 at 6:13
Paras Khosla
929214
edited Jan 17 at 6:13
Paras Khosla
929214
edited Jan 17 at 6:13
Paras Khosla
929214
929214
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
asked Jan 17 at 5:49
Tarun PrakashTarun Prakash
164
164
$begingroup$
math.stackexchange.com/questions/1531154/…
$endgroup$
– dimebucker
Jan 17 at 5:53
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
1
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
1
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
3
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37
|
show 2 more comments
$begingroup$
math.stackexchange.com/questions/1531154/…
$endgroup$
– dimebucker
Jan 17 at 5:53
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
1
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
1
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
3
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37
$begingroup$
math.stackexchange.com/questions/1531154/…
$endgroup$
– dimebucker
Jan 17 at 5:53
$begingroup$
math.stackexchange.com/questions/1531154/…
$endgroup$
– dimebucker
Jan 17 at 5:53
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
1
1
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
1
1
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
3
3
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37
|
show 2 more comments
7 Answers
7
active
oldest
votes
$begingroup$
As a check:
Partial answer, completing the square.
$x,yge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=sqrt{1/2}$.
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
An algebra-free approach:
In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=left(frac{1}{2},frac{1}{sqrt{2}}right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
$endgroup$
add a comment |
$begingroup$
As mentioned in the comments, the answer is
$$left(frac{1}{2}, frac{1}{sqrt{2}}right)$$
(or you can use $frac{sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $frac{sqrt{3}}{2}$ units, or about 0.8660 (like $sin 60^circ$).
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
$endgroup$
add a comment |
$begingroup$
We may look at differently:
Let the point we are after be $P:(x_0,sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=sqrt x$ that passes through $P$, hence its slope is $-left(frac1{2sqrt x_0}right)^{-1}$ and as such$$l_1:y=-2sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$sqrt x_0=-2sqrt x_0 (x_0-1),$$which implies $x_0=frac12$.
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
$begingroup$
The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$begin{cases}(x-1)^2+y^2=r^2,\y=sqrt xend{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$left(frac12,frac1{sqrt2}right).$$
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
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add a comment |
$begingroup$
Let $(x, sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by
$d(x)=((x-1)^2+sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.
Let $f(x):=d(x)^2.$
It is now your turn to determine $x_0 ge 0$ such that $f(x_0)= min{f(x): x ge 0 }.$
(we get $x_0 =1/2$).
We then have $d(x_0)= min{d(x): x ge 0 }.$
answered Jan 17 at 6:04
FredFred
46.9k1848
$endgroup$
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
add a comment |
$begingroup$
You can minimize the squared distance, which is
$$d^2=(x-1)^2+x=x^2-x+1=left(x-frac12right)^2+frac34.$$
Thus the minimum occurs at
$$x=frac12,$$
and
$$y=sqrt{frac12}.$$
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
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7 Answers
7
active
oldest
votes
7 Answers
7
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
As a check:
Partial answer, completing the square.
$x,yge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=sqrt{1/2}$.
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
As a check:
Partial answer, completing the square.
$x,yge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=sqrt{1/2}$.
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
add a comment |
$begingroup$
As a check:
Partial answer, completing the square.
$x,yge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=sqrt{1/2}$.
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
$endgroup$
As a check:
Partial answer, completing the square.
$x,yge 0$.
$d^2= (x-1)^2+x= $
$x^2-2x+1+x= x^2-x+1;$
$d^2=(x-1/2)^2 -1/4+1 =$
$(x-1/2)^2 +3/4ge 3/4$ (why?).
$d^2_{min} =3/4$, at $x=1/2$, $y=sqrt{1/2}$.
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
answered Jan 17 at 7:56
Peter SzilasPeter Szilas
11.4k2822
11.4k2822
add a comment |
add a comment |
$begingroup$
An algebra-free approach:
In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=left(frac{1}{2},frac{1}{sqrt{2}}right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
$endgroup$
add a comment |
$begingroup$
An algebra-free approach:
In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=left(frac{1}{2},frac{1}{sqrt{2}}right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
$endgroup$
add a comment |
$begingroup$
An algebra-free approach:
In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=left(frac{1}{2},frac{1}{sqrt{2}}right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
$endgroup$
An algebra-free approach:
In order to draw the tangent from a point $P$ on a parabola, it is sufficient to project $P$ on the axis, reflect this point with respect to the vertex and join the new point with $P$. Since the tangent drawn from $P=left(frac{1}{2},frac{1}{sqrt{2}}right)$ is orthogonal to the line joining $P$ with $(1,0)$ (by Euclid's second theorem on right triangles, or just by computing slopes), $P$ is the wanted solution.
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
answered Jan 17 at 11:40
Jack D'AurizioJack D'Aurizio
290k33282663
290k33282663
add a comment |
add a comment |
$begingroup$
As mentioned in the comments, the answer is
$$left(frac{1}{2}, frac{1}{sqrt{2}}right)$$
(or you can use $frac{sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $frac{sqrt{3}}{2}$ units, or about 0.8660 (like $sin 60^circ$).
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
$endgroup$
add a comment |
$begingroup$
As mentioned in the comments, the answer is
$$left(frac{1}{2}, frac{1}{sqrt{2}}right)$$
(or you can use $frac{sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $frac{sqrt{3}}{2}$ units, or about 0.8660 (like $sin 60^circ$).
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
$endgroup$
add a comment |
$begingroup$
As mentioned in the comments, the answer is
$$left(frac{1}{2}, frac{1}{sqrt{2}}right)$$
(or you can use $frac{sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $frac{sqrt{3}}{2}$ units, or about 0.8660 (like $sin 60^circ$).
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
$endgroup$
As mentioned in the comments, the answer is
$$left(frac{1}{2}, frac{1}{sqrt{2}}right)$$
(or you can use $frac{sqrt{2}}{2}$ for the $y$-coordinate, they're the same.) Remember that you are looking for a point on the curve, and so need to substitute the $x$-value into the function describing that curve in order to get the corresponding $y$-coordinate, and here, that means taking the square root.
The corresponding minimizing distance is, of course, $frac{sqrt{3}}{2}$ units, or about 0.8660 (like $sin 60^circ$).
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
answered Jan 17 at 7:34
The_SympathizerThe_Sympathizer
7,7102245
7,7102245
add a comment |
add a comment |
$begingroup$
We may look at differently:
Let the point we are after be $P:(x_0,sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=sqrt x$ that passes through $P$, hence its slope is $-left(frac1{2sqrt x_0}right)^{-1}$ and as such$$l_1:y=-2sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$sqrt x_0=-2sqrt x_0 (x_0-1),$$which implies $x_0=frac12$.
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
$begingroup$
We may look at differently:
Let the point we are after be $P:(x_0,sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=sqrt x$ that passes through $P$, hence its slope is $-left(frac1{2sqrt x_0}right)^{-1}$ and as such$$l_1:y=-2sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$sqrt x_0=-2sqrt x_0 (x_0-1),$$which implies $x_0=frac12$.
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
$begingroup$
We may look at differently:
Let the point we are after be $P:(x_0,sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=sqrt x$ that passes through $P$, hence its slope is $-left(frac1{2sqrt x_0}right)^{-1}$ and as such$$l_1:y=-2sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$sqrt x_0=-2sqrt x_0 (x_0-1),$$which implies $x_0=frac12$.
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
$endgroup$
We may look at differently:
Let the point we are after be $P:(x_0,sqrt x_0)$. Now say we find the equation of the line, $l_1$, that passes through $P$ and $(1,0)$. Geometrically, $l_1$ is perpendicular to a tangent line to $y=sqrt x$ that passes through $P$, hence its slope is $-left(frac1{2sqrt x_0}right)^{-1}$ and as such$$l_1:y=-2sqrt x_0 (x-1).$$
Now noting that $l_1$ also passes though $P$ we have $$sqrt x_0=-2sqrt x_0 (x_0-1),$$which implies $x_0=frac12$.
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
answered Jan 17 at 9:46
Math-funMath-fun
7,1531527
7,1531527
add a comment |
add a comment |
$begingroup$
The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$begin{cases}(x-1)^2+y^2=r^2,\y=sqrt xend{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$left(frac12,frac1{sqrt2}right).$$
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$begin{cases}(x-1)^2+y^2=r^2,\y=sqrt xend{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$left(frac12,frac1{sqrt2}right).$$
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$begin{cases}(x-1)^2+y^2=r^2,\y=sqrt xend{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$left(frac12,frac1{sqrt2}right).$$
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
$endgroup$
The solution point is such that the circle centered at $(1,0)$ tangents the curve, i.e. the system of equations
$$begin{cases}(x-1)^2+y^2=r^2,\y=sqrt xend{cases}$$ has a double root. By eliminating $y$,
$$(x-1)^2+x-r^2=0$$ has a double root when $$2(x-1)+1=0,$$ hence $$left(frac12,frac1{sqrt2}right).$$
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:42
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
$begingroup$
Let $(x, sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by
$d(x)=((x-1)^2+sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.
Let $f(x):=d(x)^2.$
It is now your turn to determine $x_0 ge 0$ such that $f(x_0)= min{f(x): x ge 0 }.$
(we get $x_0 =1/2$).
We then have $d(x_0)= min{d(x): x ge 0 }.$
answered Jan 17 at 6:04
FredFred
46.9k1848
$endgroup$
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
add a comment |
$begingroup$
Let $(x, sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by
$d(x)=((x-1)^2+sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.
Let $f(x):=d(x)^2.$
It is now your turn to determine $x_0 ge 0$ such that $f(x_0)= min{f(x): x ge 0 }.$
(we get $x_0 =1/2$).
We then have $d(x_0)= min{d(x): x ge 0 }.$
answered Jan 17 at 6:04
FredFred
46.9k1848
$endgroup$
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
add a comment |
$begingroup$
Let $(x, sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by
$d(x)=((x-1)^2+sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.
Let $f(x):=d(x)^2.$
It is now your turn to determine $x_0 ge 0$ such that $f(x_0)= min{f(x): x ge 0 }.$
(we get $x_0 =1/2$).
We then have $d(x_0)= min{d(x): x ge 0 }.$
answered Jan 17 at 6:04
FredFred
46.9k1848
$endgroup$
Let $(x, sqrt{x})$ a point on the graph. The distance to$(1,0)$ is then given by
$d(x)=((x-1)^2+sqrt{x}^2)^{1/2}=((x-1)^2+x)^{1/2}$.
Let $f(x):=d(x)^2.$
It is now your turn to determine $x_0 ge 0$ such that $f(x_0)= min{f(x): x ge 0 }.$
(we get $x_0 =1/2$).
We then have $d(x_0)= min{d(x): x ge 0 }.$
answered Jan 17 at 6:04
FredFred
46.9k1848
answered Jan 17 at 6:04
FredFred
46.9k1848
answered Jan 17 at 6:04
FredFred
46.9k1848
answered Jan 17 at 6:04
FredFred
46.9k1848
46.9k1848
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
add a comment |
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
$begingroup$
Maybe you overlooked the editing of the post. The author provided a complete solution.
$endgroup$
– A. Pongrácz
Jan 17 at 6:07
add a comment |
$begingroup$
You can minimize the squared distance, which is
$$d^2=(x-1)^2+x=x^2-x+1=left(x-frac12right)^2+frac34.$$
Thus the minimum occurs at
$$x=frac12,$$
and
$$y=sqrt{frac12}.$$
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
You can minimize the squared distance, which is
$$d^2=(x-1)^2+x=x^2-x+1=left(x-frac12right)^2+frac34.$$
Thus the minimum occurs at
$$x=frac12,$$
and
$$y=sqrt{frac12}.$$
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
$endgroup$
add a comment |
$begingroup$
You can minimize the squared distance, which is
$$d^2=(x-1)^2+x=x^2-x+1=left(x-frac12right)^2+frac34.$$
Thus the minimum occurs at
$$x=frac12,$$
and
$$y=sqrt{frac12}.$$
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
$endgroup$
You can minimize the squared distance, which is
$$d^2=(x-1)^2+x=x^2-x+1=left(x-frac12right)^2+frac34.$$
Thus the minimum occurs at
$$x=frac12,$$
and
$$y=sqrt{frac12}.$$
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
answered Jan 17 at 10:32
Yves DaoustYves Daoust
128k675227
128k675227
add a comment |
add a comment |
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math.stackexchange.com/questions/1531154/…
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– dimebucker
Jan 17 at 5:53
$begingroup$
The answer is $(1/4,1/2)$
$endgroup$
– Sauhard Sharma
Jan 17 at 5:55
1
$begingroup$
Seems good. But then the solution you obtain is $(1/2, sqrt{2}/2)$.
$endgroup$
– A. Pongrácz
Jan 17 at 6:05
1
$begingroup$
@TarunPrakash You're right. The answer is indeed $(1/2,1/sqrt{2})$. I made a small calculation error
$endgroup$
– Sauhard Sharma
Jan 17 at 6:07
3
$begingroup$
A shortcut : minimizing the distance is the same as minimizing the square of the distance.
$endgroup$
– Claude Leibovici
Jan 17 at 6:37