Mixing
Prove that every tangent of a function cuts $y$ axis at a point that is at equal distance from (0,0) and...
$begingroup$
If a function y = $frac{1}{2}$$sqrt{x-4x^2}$ is given, how would one prove that every tangent of the function cuts $y$ axis in a point that is at equal distance from point $(0, 0)$ and the point at which the tangent touches the function?Finding a derivative for every single point, constructing a tangent, finding a point where $x=0$ and then comparing given $y$ to the distance between $y$ and touching point is clearly not feasible.
real-analysis geometry derivatives
asked Jan 19 at 10:03
JoeDoughJoeDough
226
$endgroup$
add a comment |
$begingroup$
If a function y = $frac{1}{2}$$sqrt{x-4x^2}$ is given, how would one prove that every tangent of the function cuts $y$ axis in a point that is at equal distance from point $(0, 0)$ and the point at which the tangent touches the function?Finding a derivative for every single point, constructing a tangent, finding a point where $x=0$ and then comparing given $y$ to the distance between $y$ and touching point is clearly not feasible.
real-analysis geometry derivatives
asked Jan 19 at 10:03
JoeDoughJoeDough
226
$endgroup$
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12
add a comment |
$begingroup$
If a function y = $frac{1}{2}$$sqrt{x-4x^2}$ is given, how would one prove that every tangent of the function cuts $y$ axis in a point that is at equal distance from point $(0, 0)$ and the point at which the tangent touches the function?Finding a derivative for every single point, constructing a tangent, finding a point where $x=0$ and then comparing given $y$ to the distance between $y$ and touching point is clearly not feasible.
real-analysis geometry derivatives
asked Jan 19 at 10:03
JoeDoughJoeDough
226
$endgroup$
If a function y = $frac{1}{2}$$sqrt{x-4x^2}$ is given, how would one prove that every tangent of the function cuts $y$ axis in a point that is at equal distance from point $(0, 0)$ and the point at which the tangent touches the function?Finding a derivative for every single point, constructing a tangent, finding a point where $x=0$ and then comparing given $y$ to the distance between $y$ and touching point is clearly not feasible.
real-analysis geometry derivatives
real-analysis geometry derivatives
asked Jan 19 at 10:03
JoeDoughJoeDough
226
asked Jan 19 at 10:03
JoeDoughJoeDough
226
asked Jan 19 at 10:03
JoeDoughJoeDough
226
asked Jan 19 at 10:03
JoeDoughJoeDough
226
asked Jan 19 at 10:03
JoeDoughJoeDough
226
226
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12
add a comment |
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:
$$y_0^2=frac 14(x_0-4x_0^2)tag{1}$$
Equation of tangent passing through $T$ is:
$$(y-y_0)=y'_0(x-x_0)$$
where:
$$y'_0=frac{1-8x_0}{4sqrt{x_0-4x_0^2}}=frac{1-8x_0}{8y_0}$$
Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:
$$y_1-y_0=frac{8x_0^2-x_0}{8y_0}tag{2}$$
You have to show that: $TP=OP$ or $TP^2=OP^2$:
$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$
$$(y_1-y_0)^2+x_0^2=y_1^2$$
$$-2y_0y_1+y_0^2+x_0^2=0$$
$$x_0^2+y_0^2=2y_0y_1tag{3}$$
Combine (2) and (3) and you get:
$$x_0^2+y_0^2=2y_0(y_0+frac{8x_0^2-x_0}{8y_0})$$
$$x_0^2=y_0^2+frac14(8x_0^2-x_0)$$
$$4x_0^2=4y_0^2+8x_0^2-x_0$$
$$x_0-4x_0^2=4y_0^2$$
...which is always true, according to (1)
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
$endgroup$
add a comment |
$begingroup$
The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).
So, tra la, you want to show
$p = sqrt{(p - y(a))^2 + a^2}.$
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
$begingroup$
This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
$endgroup$
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%2f3079176%2fprove-that-every-tangent-of-a-function-cuts-y-axis-at-a-point-that-is-at-equal%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
$begingroup$
Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:
$$y_0^2=frac 14(x_0-4x_0^2)tag{1}$$
Equation of tangent passing through $T$ is:
$$(y-y_0)=y'_0(x-x_0)$$
where:
$$y'_0=frac{1-8x_0}{4sqrt{x_0-4x_0^2}}=frac{1-8x_0}{8y_0}$$
Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:
$$y_1-y_0=frac{8x_0^2-x_0}{8y_0}tag{2}$$
You have to show that: $TP=OP$ or $TP^2=OP^2$:
$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$
$$(y_1-y_0)^2+x_0^2=y_1^2$$
$$-2y_0y_1+y_0^2+x_0^2=0$$
$$x_0^2+y_0^2=2y_0y_1tag{3}$$
Combine (2) and (3) and you get:
$$x_0^2+y_0^2=2y_0(y_0+frac{8x_0^2-x_0}{8y_0})$$
$$x_0^2=y_0^2+frac14(8x_0^2-x_0)$$
$$4x_0^2=4y_0^2+8x_0^2-x_0$$
$$x_0-4x_0^2=4y_0^2$$
...which is always true, according to (1)
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
$endgroup$
add a comment |
$begingroup$
Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:
$$y_0^2=frac 14(x_0-4x_0^2)tag{1}$$
Equation of tangent passing through $T$ is:
$$(y-y_0)=y'_0(x-x_0)$$
where:
$$y'_0=frac{1-8x_0}{4sqrt{x_0-4x_0^2}}=frac{1-8x_0}{8y_0}$$
Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:
$$y_1-y_0=frac{8x_0^2-x_0}{8y_0}tag{2}$$
You have to show that: $TP=OP$ or $TP^2=OP^2$:
$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$
$$(y_1-y_0)^2+x_0^2=y_1^2$$
$$-2y_0y_1+y_0^2+x_0^2=0$$
$$x_0^2+y_0^2=2y_0y_1tag{3}$$
Combine (2) and (3) and you get:
$$x_0^2+y_0^2=2y_0(y_0+frac{8x_0^2-x_0}{8y_0})$$
$$x_0^2=y_0^2+frac14(8x_0^2-x_0)$$
$$4x_0^2=4y_0^2+8x_0^2-x_0$$
$$x_0-4x_0^2=4y_0^2$$
...which is always true, according to (1)
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
$endgroup$
add a comment |
$begingroup$
Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:
$$y_0^2=frac 14(x_0-4x_0^2)tag{1}$$
Equation of tangent passing through $T$ is:
$$(y-y_0)=y'_0(x-x_0)$$
where:
$$y'_0=frac{1-8x_0}{4sqrt{x_0-4x_0^2}}=frac{1-8x_0}{8y_0}$$
Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:
$$y_1-y_0=frac{8x_0^2-x_0}{8y_0}tag{2}$$
You have to show that: $TP=OP$ or $TP^2=OP^2$:
$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$
$$(y_1-y_0)^2+x_0^2=y_1^2$$
$$-2y_0y_1+y_0^2+x_0^2=0$$
$$x_0^2+y_0^2=2y_0y_1tag{3}$$
Combine (2) and (3) and you get:
$$x_0^2+y_0^2=2y_0(y_0+frac{8x_0^2-x_0}{8y_0})$$
$$x_0^2=y_0^2+frac14(8x_0^2-x_0)$$
$$4x_0^2=4y_0^2+8x_0^2-x_0$$
$$x_0-4x_0^2=4y_0^2$$
...which is always true, according to (1)
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
$endgroup$
Denote point where tangent $t$ touches graph of the function with $T(x_0, y_0)$. Obviously:
$$y_0^2=frac 14(x_0-4x_0^2)tag{1}$$
Equation of tangent passing through $T$ is:
$$(y-y_0)=y'_0(x-x_0)$$
where:
$$y'_0=frac{1-8x_0}{4sqrt{x_0-4x_0^2}}=frac{1-8x_0}{8y_0}$$
Tangent intersect $y$ axis at point $P_(x_1=0, y_1)$. If you replace these cooridinates into the equaton of the tangent:
$$y_1-y_0=frac{8x_0^2-x_0}{8y_0}tag{2}$$
You have to show that: $TP=OP$ or $TP^2=OP^2$:
$$TP^2=(x_1-x_0)^2+(y_1-y_0)^2=OP^2=y_1^2$$
$$(y_1-y_0)^2+x_0^2=y_1^2$$
$$-2y_0y_1+y_0^2+x_0^2=0$$
$$x_0^2+y_0^2=2y_0y_1tag{3}$$
Combine (2) and (3) and you get:
$$x_0^2+y_0^2=2y_0(y_0+frac{8x_0^2-x_0}{8y_0})$$
$$x_0^2=y_0^2+frac14(8x_0^2-x_0)$$
$$4x_0^2=4y_0^2+8x_0^2-x_0$$
$$x_0-4x_0^2=4y_0^2$$
...which is always true, according to (1)
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
answered Jan 19 at 10:50
OldboyOldboy
8,59011036
8,59011036
add a comment |
add a comment |
$begingroup$
The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).
So, tra la, you want to show
$p = sqrt{(p - y(a))^2 + a^2}.$
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
$begingroup$
The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).
So, tra la, you want to show
$p = sqrt{(p - y(a))^2 + a^2}.$
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
$endgroup$
add a comment |
$begingroup$
The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).
So, tra la, you want to show
$p = sqrt{(p - y(a))^2 + a^2}.$
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
$endgroup$
The equation of for the tangent at x = a is
y - y(a) = y'(a)(x - a).
When x = 0, the y intercept is p = y(a) - ay'(a).
So, tra la, you want to show
$p = sqrt{(p - y(a))^2 + a^2}.$
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
answered Jan 19 at 11:01
William ElliotWilliam Elliot
8,3122720
8,3122720
add a comment |
add a comment |
$begingroup$
This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
$endgroup$
add a comment |
$begingroup$
This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
$endgroup$
add a comment |
$begingroup$
This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
$endgroup$
This is the equation of a half-circle passing through $(0,0)$ and tangent there to the $y$ axis. The property to prove can be then rephrased as follows: prove that the two segments of tangent, issued from any point on the positive $y$ axis, are equal, which is a well known geometric feature of any circle.
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
answered Jan 19 at 14:52
AretinoAretino
24.5k21444
24.5k21444
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%2f3079176%2fprove-that-every-tangent-of-a-function-cuts-y-axis-at-a-point-that-is-at-equal%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
$begingroup$
Why can't it be done?
$endgroup$
– William Elliot
Jan 19 at 10:32
$begingroup$
Because there are infinitely many points on x axis?
$endgroup$
– JoeDough
Jan 19 at 10:41
$begingroup$
For what reason?
$endgroup$
– JoeDough
Jan 19 at 10:50
$begingroup$
You have received 3 answers, consider accepting one of them
$endgroup$
– Oldboy
Jan 21 at 9:12