Find the direction in which slope is the steepest

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Problem:

The height of a hill (in meters) is given by [z=(2xy)-(3x^2)-(4y^2)-(18x)+(28y)+12], where x is the distance east, y is the distance north of the origin. In which compass direction is the slope at x=y=1 the steepest?

Solution (so far):

Does it have something to do with the gradient?

Does it have something to do with the gradient?
Since the gradient vector points in maximal increase of a multivariate function, and z here represents a height then grad(z) will point where the height increases most rapidly, I.e where it is most steep. So yes, the gradient would certainly be helpful here.

I thought so, but I don't see how I'd be able to find the direction by calculating grad(z). Any suggestions?

I thought so, but I don't see how I'd be able to find the direction by calculating grad(z). Any suggestions?

The ##∇z## will give you a vector ...

So, the gradient of Z at a point is a vector pointing in the direction of the steepest slope.

I calculated ## \nabla z(x,y)= (-6x+2y-18)\hat{i}+(2x-8y+28)\hat{j} ##.

So this is a vector pointing in the direction of greatest increase (ie. the steepest).

Thus, ## \nabla z(1,1)= (-6+2-18)\hat{i}+(2-8+28)\hat{j}=(-6+2-18)\hat{i}+(2-8+28)\hat{j}=-22\hat{i}+22\hat{j} ##

The ##∇z## will give you a vector ...

So, the gradient of Z at a point is a vector pointing in the direction of the steepest slope.

Thanks. I calculated the gradient. I'm not sure what to do next.

I found ## |\nabla z(1,1)|= 22 \sqrt{2} ##. So, by the dot product, the angle this vector makes with the x-axis is given by ## cos \theta = \frac{\nabla z(1,1) \cdot \hat{x}}{|\nabla z(1,1)| \cdot |\hat{x}|} = \frac{-22}{22\sqrt{2} \cdot 1 }= \frac{1}{\sqrt{2}} \Rightarrow \theta = 135°##.

I think I messed up somewhere because my text is saying correct answer is ## \theta = -45°##.

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I think I messed up somewhere because my is saying correct answer is ## \theta = -45°##.

Hum.. Perhaps your initial function lacks a minus somewhere?

Well, the steepest ascent is also associated with the steepest descent... just the opposite direction (180° turn).

gneil, I understand what you mean, but in that case shouldn't -45° & 135° both be acceptable answers?

gneil, I understand what you mean, but in that case shouldn't -45° & 135° both be acceptable answers?

Yes, I would think so.