Geometric interpretation of differential results
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I had a question which reads as follows.
A triangle has two of its angular points at $(a,0)$ and $(b,0)$, and the third point $(x,y)$ is movable along $y=x$. If $A$ is an area of the triangle; show that $2(frac{dA}{dx})=a+b$
While showing the result is no big deal, how do i think about this proposition geometrically? Or rather how do i approach the geometric aspect of these types of 'rates of changes' in general
calculus derivatives
edited 2 days ago
S. O.
684
asked 2 days ago
user182947
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show 2 more comments
up vote
1
down vote
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I had a question which reads as follows.
A triangle has two of its angular points at $(a,0)$ and $(b,0)$, and the third point $(x,y)$ is movable along $y=x$. If $A$ is an area of the triangle; show that $2(frac{dA}{dx})=a+b$
While showing the result is no big deal, how do i think about this proposition geometrically? Or rather how do i approach the geometric aspect of these types of 'rates of changes' in general
calculus derivatives
edited 2 days ago
S. O.
684
asked 2 days ago
user182947
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday
|
show 2 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I had a question which reads as follows.
A triangle has two of its angular points at $(a,0)$ and $(b,0)$, and the third point $(x,y)$ is movable along $y=x$. If $A$ is an area of the triangle; show that $2(frac{dA}{dx})=a+b$
While showing the result is no big deal, how do i think about this proposition geometrically? Or rather how do i approach the geometric aspect of these types of 'rates of changes' in general
calculus derivatives
edited 2 days ago
S. O.
684
asked 2 days ago
user182947
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
I had a question which reads as follows.
A triangle has two of its angular points at $(a,0)$ and $(b,0)$, and the third point $(x,y)$ is movable along $y=x$. If $A$ is an area of the triangle; show that $2(frac{dA}{dx})=a+b$
While showing the result is no big deal, how do i think about this proposition geometrically? Or rather how do i approach the geometric aspect of these types of 'rates of changes' in general
calculus derivatives
calculus derivatives
edited 2 days ago
S. O.
684
asked 2 days ago
user182947
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
S. O.
684
asked 2 days ago
user182947
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
S. O.
684
edited 2 days ago
S. O.
684
edited 2 days ago
S. O.
684
684
asked 2 days ago
user182947
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
user182947
183
asked 2 days ago
user182947
183
183
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
user182947 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday
|
show 2 more comments
$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday
$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday
|
show 2 more comments
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user182947 is a new contributor. Be nice, and check out our Code of Conduct.
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$A$ is an area or the area of the triangle? If 'the', then assume $b > a$, therefore $A(x) = (b-a)x$ and $frac{dA}{dx} = b-a$. I'm kinda confused there...
– Christopher Marley
2 days ago
I believe its 'The' but i used the wordings of the question. I think its safe to assume it'd be The
– user182947
2 days ago
The original version of the question says 'the', so I'll go with that. Then we would have $2(frac{dA}{dx}) = b-a$, instead of $ = b+a$.
– Christopher Marley
2 days ago
If a = b @ChristopherMarley, then the area would be static.
– William Elliot
2 days ago
Assume $ane b.$The length of the base of the triangle is $|a-b|.$ The height of $(x,y)=(x,x)$ from the base-line $Bbb Rtimes {0}$ is $|y|=|x|$ So $A=|x|cdot |a-b|/2.$ If $x>0$ then $dA/dx=|a-b|/2.$
– DanielWainfleet
yesterday