how do u find the gradient of y-3x=2

in terms of calculus, the gradient is defined to be a vector field, that is, given a function it will assign a vector to each point of the function. the components of each vector tell how much the function is changing in that direction.

$$grad(f) = \frac{\partial{f}}{\partial{x}}i + \frac{\partial{f}}{\partial{y}}j$$

is the vector field. so in your example:

$$\frac{\partial{f}}{\partial{x}} = -3$$

$$\frac{\partial{f}}{\partial{y}} = 1$$

so:

$$grad(y-3x-2) = -3i + j$$

notice that the component of the gradient of your function are constant...thats because your function just a line. if your function were something more complicated, then your components would be functions and your would evaluate them at a particular point because your gradient would then vary as a true vector field.

mathwonk
Homework Helper
i am puzzled. you have not said what the function is, so i do not know what the rgadient is.

if the function is f(x,y) = y-3x = -3x+y, then the gradient is the same everywhere, namely (-3,1).

same if the function is f(x,y) = -3x+y-2 as has been assumed above, but this is not clear from your question. an equation is not a function, unless meant sas the graph of the function, in which case you would be giving the function y = 2-3x whose "gradient is -3.

I think he just wanted the gradient of a straight line..

mathwonk said:
i am puzzled. you have not said what the function is, so i do not know what the rgadient is.

if the function is f(x,y) = y-3x = -3x+y, then the gradient is the same everywhere, namely (-3,1).

same if the function is f(x,y) = -3x+y-2 as has been assumed above, but this is not clear from your question. an equation is not a function, unless meant sas the graph of the function, in which case you would be giving the function y = 2-3x whose "gradient is -3.
The gradient vector isnt perpendicular to the slope?

whozum said:
The gradient vector isnt perpendicular to the slope?
No, for a linear function the gradient is the slope (in "British-speaking" places).

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