How Do You Find the Gradient of a Line?

[SleSSi]

i need to find the gradient of the line y-3x=2 how do i do it again?

 

[Nylex]

Put it in the form y = mx + c and m is the gradient.

 

[dextercioby]

What's the connection between the slope of the line & its gradient...?

Daniel.

 

[BenGoodchild]

hey,

{y-displacement}/{x-displacement} = gradient displacement is how much y changes and x changes from two points.

The gradient IS the slope of the line


Regards,

Ben

 

[futb0l]

Rise over Run - is how I remember it.

 

[whozum]

Wait, is this gradient the same gradient as in gradient of a vector field or does it mean gradient like slope of the line?

 

[BenGoodchild]

Wait, is this gradient the same gradient as in gradient of a vector field or does it mean gradient like slope of the line?

Slope of line - y=mx +c

Regards,

Ben

 

[HallsofIvy]

The original question was about y- 3x= 2, a real valued linear function of a single real variable. I'm afraid all that talk about "vectors" and even "derivatives" will just confuse the original poster.

Slessi: solve for y. In this example, solving for y gives y= 3x+ 2 so the gradient (slope) is 3. (Almost) any linear equation can be solved for y in the form y= mx+ b and the gradient is the number m.

(I say "(Almost)" because a vertical straight line, like x= 1, cannot be put in that form: it has NO gradient.

 

[BenGoodchild]

(I say "(Almost)" because a vertical straight line, like x= 1, cannot be put in that form: it has NO gradient.

I was just staring at this statement whilst drinking tea and began to think, there is a gradient because gradient is [y-displacement]/[x-displacement] and in this case, 0/change in x. Change in x is always a non-zero integer value and therefore the gradient is 0/non-zero integer = 0. Looking at the graph, this fits, the gradient looks to be 0.

However, if the graph is of y=1 ,then the equation becomes: [change in y]/0. Change in y is always a non-zero integer value and therefore the gradient is non-zero integer/0 which is mathematically undefined. However, if one looks at the graph in a topolgical sense (Pemrose is my source on the idea of topolgy), then the gradient looks to be -infinity or +infinity. And this leads one to consider what result dividing by 0 gives.

Anyway, those are just my thoughts,


Regards,

Ben

 

[HallsofIvy]

[y-displacement]/[x-displacement]

Yes, that's exactly right.

and in this case, 0/change in x.

No, that's exactly wrong. The equation in question is x= 1. x is always 1: x doesn't change, y can be anything: the equation is change in y/0.

. Change in x is always a non-zero integer value

Where in the world did you get that idea? If x= 1 the change is 0! x= 1 means exactly that: x is always 1, not the "change" in x!

However, if the graph is of y=1 ,then the equation becomes: [change in y]/0.

Same error: if y= 1 then y does not chage: the slope is 0/change in x= 0.

 

[BenGoodchild]

Hi, I commented on x=1 instead of y=1 and visa versa - please reread the post in its now corrected form (below) and give your throughts.

I was just staring at this statement whilst drinking tea and began to think, there is a gradient because gradient is [y-displacement]/[x-displacement] and in the case of y=1, 0/change in x. Change in x is always a non-zero integer value and therefore the gradient is 0/non-zero integer = 0. Looking at the graph, this fits, the gradient looks to be 0.

However, if the graph is of x=1 ,then the equation becomes: [change in y]/0. Change in y is always a non-zero integer value and therefore the gradient is non-zero integer/0 which is mathematically undefined. However, if one looks at the graph in a topolgical sense (Pemrose is my source on the idea of topolgy), then the gradient looks to be -infinity or +infinity. And this leads one to consider what result dividing by 0 gives.

-Ben