Hi,

I am stuck on the second and third part of this question, to be honest I just have no idea how to draw contour lines! I know how stationary points look on contour maps but that’s pretty much it,

How would I go about drawing contour lines here? Also I attempted the third part but no clue if that’s what you are supposed to do, below are the question and my attempt

I am stuck on the second and third part of this question, to be honest I just have no idea how to draw contour lines! I know how stationary points look on contour maps but that’s pretty much it,

How would I go about drawing contour lines here? Also I attempted the third part but no clue if that’s what you are supposed to do, below are the question and my attempt

(Throwing some ideas here, not sure how useful they are. Curve sketching is really all about noticing features and piecing every bits of clues together)

- The function is symmetric along x- and y-axes (clue: the function only has even powers of x's and y's), so looking at one quadrant will do. (I don't dare saying it's "even" for multivariate functions, as it doesn't quite make sense, but that's the idea anyway)

- If you rearrange terms, following presumably standard procedures for sketching contour lines (IDK what they are, really), you should get y is a quadratic in x^2, with some sort of vertical shifting controlled by what the function value takes. So I would expect some sort of quartic-looking curves.

- The x^2 term gives the "dimples" look to the quartic curves (this feels like a "experimental result from desmos").

- Stationary points are useful, obviously. Contour lines taken at different function values should "tend towards/away" stationary points, if that makes sense.

- The function is symmetric along x- and y-axes (clue: the function only has even powers of x's and y's), so looking at one quadrant will do. (I don't dare saying it's "even" for multivariate functions, as it doesn't quite make sense, but that's the idea anyway)

- If you rearrange terms, following presumably standard procedures for sketching contour lines (IDK what they are, really), you should get y is a quadratic in x^2, with some sort of vertical shifting controlled by what the function value takes. So I would expect some sort of quartic-looking curves.

- The x^2 term gives the "dimples" look to the quartic curves (this feels like a "experimental result from desmos").

- Stationary points are useful, obviously. Contour lines taken at different function values should "tend towards/away" stationary points, if that makes sense.

(edited 1 year ago)

I would check on your calculation for $f_{xx}$.

For ii), when a = 1, the graph is clearly symmetric about the x and y axis. However, as there is an x^4, this quickly outgrows y^2 term when greater than 1. Therefore, you would draw shapes that gradually get more oblong [due to the quartic term], and also stretch more in the y axis (as a larger y value needs to compensate for a small decrease in the x value). Furthermore, as there is only one SP which is a minima you know the contours expand radially from the origin.

Unfortunately the case when a = -1 is far more complex and fiddly to draw.

First of all, you have saddles at (1, 0) and (-1, 0), and a minima at (0, 0). Therefore, you can deduce that at say (1 + k, 0), the value of f would be less than (1, 0). Furthermore, you can also deduce that there must be a contour between (1, 0) and (-1, 0) at the same value as (1+k, 0), hence you end up with 2 separate contour lines for certain values.

You could think of the contours of x^2 + y^2 - x^4/2 as they are equivalent, but other than that I cannot provide any more insight. Maybe graph in say desmos, but I personally would have never gotten this part in an exam. It is far too complex for me to provide an accurate sketch of.

For iii), We want a contour that passes through (1, 0) and (-1, 0). Because the function inside the exponential is even, if we were to find such a contour for one point it would contain the other. The contour must have the value of f(1, 0), which is e^1/4, hence our contour is $exp(...) = e^{\frac{1}{4}}$. Then, take the log of both sides and solve for y.

Finally, no contour passes through (0, 0), as e^x is never equal to 0.

As a final point, you should include the z value for the SP's, i.e. (0, 0, 1) not (0, 0), which may be pedantic but it could lose you marks if you don't.

For ii), when a = 1, the graph is clearly symmetric about the x and y axis. However, as there is an x^4, this quickly outgrows y^2 term when greater than 1. Therefore, you would draw shapes that gradually get more oblong [due to the quartic term], and also stretch more in the y axis (as a larger y value needs to compensate for a small decrease in the x value). Furthermore, as there is only one SP which is a minima you know the contours expand radially from the origin.

Unfortunately the case when a = -1 is far more complex and fiddly to draw.

First of all, you have saddles at (1, 0) and (-1, 0), and a minima at (0, 0). Therefore, you can deduce that at say (1 + k, 0), the value of f would be less than (1, 0). Furthermore, you can also deduce that there must be a contour between (1, 0) and (-1, 0) at the same value as (1+k, 0), hence you end up with 2 separate contour lines for certain values.

You could think of the contours of x^2 + y^2 - x^4/2 as they are equivalent, but other than that I cannot provide any more insight. Maybe graph in say desmos, but I personally would have never gotten this part in an exam. It is far too complex for me to provide an accurate sketch of.

For iii), We want a contour that passes through (1, 0) and (-1, 0). Because the function inside the exponential is even, if we were to find such a contour for one point it would contain the other. The contour must have the value of f(1, 0), which is e^1/4, hence our contour is $exp(...) = e^{\frac{1}{4}}$. Then, take the log of both sides and solve for y.

Finally, no contour passes through (0, 0), as e^x is never equal to 0.

As a final point, you should include the z value for the SP's, i.e. (0, 0, 1) not (0, 0), which may be pedantic but it could lose you marks if you don't.

(edited 1 year ago)

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