A circle of radius $r$ is dropped into the parabola $y=x^{2}$. Find the largest $r$ so the circle will touch...
$begingroup$
If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
calculus geometry functions
asked Aug 19 '15 at 9:27
user3063381user3063381
446
$endgroup$
add a comment |
$begingroup$
If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
calculus geometry functions
asked Aug 19 '15 at 9:27
user3063381user3063381
446
$endgroup$
2
$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29
$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41
add a comment |
$begingroup$
If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
calculus geometry functions
asked Aug 19 '15 at 9:27
user3063381user3063381
446
$endgroup$
If $r$ is too large, the circle will not fall to the bottom, if $r$ is sufficiently small, the circle will touch the parabola at its vertex $(0, 0)$. Find the largest value of $r$ s.t. the circle will touch the vertex of the parabola.
calculus geometry functions
calculus geometry functions
asked Aug 19 '15 at 9:27
user3063381user3063381
446
asked Aug 19 '15 at 9:27
user3063381user3063381
446
asked Aug 19 '15 at 9:27
user3063381user3063381
446
asked Aug 19 '15 at 9:27
user3063381user3063381
446
asked Aug 19 '15 at 9:27
user3063381user3063381
446
446
2
$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29
$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41
add a comment |
2
$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29
$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41
2
2
$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29
$begingroup$
That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:29
$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41
$begingroup$
I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
$endgroup$
– user3063381
Aug 19 '15 at 9:41
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
$endgroup$
add a comment |
$begingroup$
Following my comment, you just have to compute for which $r$s
$$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=frac{1}{2}$.
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
add a comment |
$begingroup$
Maximum radius is also the curvature at x=0.
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
$endgroup$
add a comment |
$begingroup$
Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 Rightarrow x_1=0;\
x^2=2r-1 le 0 Rightarrow rle frac12.$$
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
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$begingroup$
Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
$endgroup$
add a comment |
$begingroup$
Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
$endgroup$
add a comment |
$begingroup$
Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
$endgroup$
Let the center of the circle $(0, r)$ & radius $r$ hence the equation of the circle $$(x-0)^2+(y-r)^2=r^2$$ $$x^2+(y-r)^2=r^2$$ Now, solving the equations of parabola $y=x^2$ & equation of the circle we get $$x^2+(x^2-r)^2=r^2$$ $$x^2+x^4+r^2-2rx^2=r^2$$ $$x^4-(2r-1)x^2=0$$ $$x^2(x^2+1-2r)=0$$
$$x^2=0 vee x^2+1-2r=0$$ $$x=0 vee x^2=2r-1$$
$$x=0 vee x^=pmsqrt{2r-1}$$ Corresponding values of $y$ are calculated as follows
$$(x=0, y=0), (x=sqrt{2r-1}, y=2r-1), (x=-sqrt{2r-1}, y=2r-1)$$
Thus, we find the three points of intersection of the circle with the parabola $(0, 0)$, $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$
But the circle is touching the parabola at the vertex $(0, 0)$ only hence, it is possible when other two points $(sqrt{2r-1}, 2r-1)$ & $(-sqrt{2r-1}, 2r-1)$ are coinciding with the vertex $(0, 0)$ hence, we get $$pm sqrt{2r-1}=0iff r=frac{1}{2}$$
& $$2r-1=0iff r=frac{1}{2}$$
$$bbox[5px, border:2px solid #C0A000]{color{red}{text{Maximum radius of circle, }r_{text{max}}=color{blue}{frac{1}{2}}}}$$
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
edited Aug 19 '15 at 10:06
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
answered Aug 19 '15 at 9:53
Harish Chandra RajpootHarish Chandra Rajpoot
29.7k103772
29.7k103772
add a comment |
add a comment |
$begingroup$
Following my comment, you just have to compute for which $r$s
$$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=frac{1}{2}$.
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
add a comment |
$begingroup$
Following my comment, you just have to compute for which $r$s
$$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=frac{1}{2}$.
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
add a comment |
$begingroup$
Following my comment, you just have to compute for which $r$s
$$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=frac{1}{2}$.
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
$endgroup$
Following my comment, you just have to compute for which $r$s
$$ f(x)=sqrt{r^2-x^2} = r-x^2 = g(x) $$
just at $x=0$. Since:
$$ (r^2-x^2)-(r-x^2)^2 = x^2(2r-1-x^2) $$
the critical radius is obviously $r=frac{1}{2}$.
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
answered Aug 19 '15 at 9:35
Jack D'AurizioJack D'Aurizio
292k33284672
292k33284672
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
add a comment |
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
Could you explain the first equation?
$endgroup$
– user3063381
Aug 19 '15 at 9:44
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
$begingroup$
@user3063381: $f$ and $g$ represent the semicircle having center in the origin through $(0,r)$ and the parabola $y=x^2$ reflected with respect to $(0,r/2)$. So, instead of dropping a circle inside a parabola, I dropped a parabola with fixed $a$ on a circle with radius $r$.
$endgroup$
– Jack D'Aurizio
Aug 19 '15 at 9:47
add a comment |
$begingroup$
Maximum radius is also the curvature at x=0.
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
$endgroup$
add a comment |
$begingroup$
Maximum radius is also the curvature at x=0.
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
$endgroup$
add a comment |
$begingroup$
Maximum radius is also the curvature at x=0.
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
$endgroup$
Maximum radius is also the curvature at x=0.
Curvature of a plane curve is d/dx(dy/dx)/(1 + (dy/dx)^2)^ 3/2
For Parabola y = x^2,
The value of curvature is 2. Therefore R = 1/2
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
answered Dec 24 '18 at 23:21
Venkatesan GopalakrishnanVenkatesan Gopalakrishnan
111
111
add a comment |
add a comment |
$begingroup$
Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 Rightarrow x_1=0;\
x^2=2r-1 le 0 Rightarrow rle frac12.$$
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
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Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 Rightarrow x_1=0;\
x^2=2r-1 le 0 Rightarrow rle frac12.$$
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
$endgroup$
add a comment |
$begingroup$
Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 Rightarrow x_1=0;\
x^2=2r-1 le 0 Rightarrow rle frac12.$$
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
$endgroup$
Idea is similar with the accepted answer, yet it is simpler.
The equation of the circle with center $(0,r)$ is: $(x-0)^2+(y-r)^2=r^2$.
The equation of the parabola is: $y=x^2$.
The two graphs must have a single point of contact at: $(0,0)$.
Substitute $y=x^2$ into the equation of circle:
$$y+(y-r)^2=r^2 Rightarrow y^2+(1-2r)y=0 Rightarrow y_1=0; y_2=2r-1.$$
We find:
$$x^2=0 Rightarrow x_1=0;\
x^2=2r-1 le 0 Rightarrow rle frac12.$$
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
answered Dec 25 '18 at 5:56
farruhotafarruhota
21.8k2842
21.8k2842
add a comment |
add a comment |
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That is equivalent to finding the osculating circle at the vertex of the parabola, i.e. to computing a curvature. What have you attempted?
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– Jack D'Aurizio
Aug 19 '15 at 9:29
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I thought that I would need to somehow use the equation of the circle to find the circle that intersects the parabola at the origin while pointing up the $y$ direction. That is $(x -0)^{2} - (y-h)^{2} = r^{2}$. I know that $x$ must be at $0$ but I don't know that value of $y$ nor the radius.
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– user3063381
Aug 19 '15 at 9:41