Tangent points of a circle touching two functions
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I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!
See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.
If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.
Thank you!
BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.
calculus
New contributor
add a comment |
up vote
1
down vote
favorite
I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!
See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.
If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.
Thank you!
BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.
calculus
New contributor
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!
See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.
If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.
Thank you!
BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.
calculus
New contributor
I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!
See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.
If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.
Thank you!
BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.
calculus
calculus
New contributor
New contributor
edited 18 hours ago
Robert Z
89.3k1056128
89.3k1056128
New contributor
asked 19 hours ago
Sejin Kim
61
61
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New contributor
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add a comment |
3 Answers
3
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oldest
votes
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0
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HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
add a comment |
up vote
0
down vote
HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:quad (x_1, log ,x_1) $$
Slope is the derivative evaluated at that point.
$$ tan phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 cot phi $$
Coordinates $$ R: (0,, log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$
add a comment |
up vote
0
down vote
Start from the end and go back.
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.
Can you take it from here?
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
add a comment |
up vote
0
down vote
HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
add a comment |
up vote
0
down vote
up vote
0
down vote
HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
HINT:
I think the equations should be
$$y=ln(x)........(1)$$
$$y=ln(-x)........(2)$$
$$(x-h)^2+(y-k)^2=4........(3)$$
Now at tangent point the slope of the circle and the given graph will be the same.so,
$$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$
You will get a system of equations here..4 equations,4 unknown...
Just solve them.
answered 18 hours ago
Rakibul Islam Prince
80529
80529
add a comment |
add a comment |
up vote
0
down vote
HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:quad (x_1, log ,x_1) $$
Slope is the derivative evaluated at that point.
$$ tan phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 cot phi $$
Coordinates $$ R: (0,, log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$
add a comment |
up vote
0
down vote
HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:quad (x_1, log ,x_1) $$
Slope is the derivative evaluated at that point.
$$ tan phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 cot phi $$
Coordinates $$ R: (0,, log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$
add a comment |
up vote
0
down vote
up vote
0
down vote
HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:quad (x_1, log ,x_1) $$
Slope is the derivative evaluated at that point.
$$ tan phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 cot phi $$
Coordinates $$ R: (0,, log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$
HINTS:
Draw normal/tangent at $Q$ cutting symmetry line at $R$
Coordinates $$ Q:quad (x_1, log ,x_1) $$
Slope is the derivative evaluated at that point.
$$ tan phi= 1/x =1/x_1 $$
Difference along central symmetry axis, vertical component of $QR$
$$ h = x_1 cot phi $$
Coordinates $$ R: (0,, log x_1 +h) $$
Equation of circle tangent at $ (P,Q) $ centered at $R:$
$$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$
edited 17 hours ago
answered 18 hours ago
Narasimham
20.4k52158
20.4k52158
add a comment |
add a comment |
up vote
0
down vote
Start from the end and go back.
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.
Can you take it from here?
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
add a comment |
up vote
0
down vote
Start from the end and go back.
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.
Can you take it from here?
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
add a comment |
up vote
0
down vote
up vote
0
down vote
Start from the end and go back.
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.
Can you take it from here?
Start from the end and go back.
Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
The final step is to find $x_0>0$ such that $|RQ|=2$.
Can you take it from here?
edited 17 hours ago
answered 19 hours ago
Robert Z
89.3k1056128
89.3k1056128
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
add a comment |
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
Is there a way to find the generic x-value algebraically, rather than graphically?
– Sejin Kim
18 mins ago
add a comment |
Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.
Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.
Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.
Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.
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