How do I trace the contour lines ` z=k` for` z=e^(y-x^2)` when `k=1, e^2, e^-1, e^-2.`I'm confused mostly about how to isolate the x and y values. I know I have to make z=k<=>e(y-x^2)=k and...

How do I trace the contour lines ` z=k` for` z=e^(y-x^2)` when `k=1, e^2, e^-1, e^-2.`

I'm confused mostly about how to isolate the x and y values. I know I have to make z=k<=>e(y-x^2)=k and make it something like y-x^2=e^k?
I'm soo confusedd! Thank you for helping, in advance!

Asked on by sujames

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txmedteach | High School Teacher | (Level 3) Associate Educator

Posted on

The contour lines pretty much describe cross sections of the surface, like a topographic map. To get these, we just set z=k and solve for y in terms of x.

Let's see if we can get a general form first:

`z = e^(y-x^2)`

Taking the natural log of both sides:

`lnz = y-x^2`

Now, we just add `x^2` t0 both sides to isolate y:

`y = x^2 + lnz`

Now, when we want a contour line, we just substitute the above values for z:

1) k = 1

`y = x^2 + ln 1 = x^2`

Our contour line in this case is simply `y=x^2`.

2) `k = e^2`

`y = x^2 + lne^2 = x^2 + 2`

This contour line is simply the same as the one above, just shifted up by 2.

3) `k = e^-1`

Same thing:

`y = x^2+lne^-1 = x^2-1`

Again, same as the first case, just shifted down by one.

4) `k=e^-2`

`y = x^2 +lne^-2 = x^2-2`

Again, same as first case, but shifted down by two this time.

Let's show these on a graph to get a sense of what we're looking at:

For `k = 1` we'll make the curve black.

For `k = e^2` we'll make the curve green.

For `k = e^-1` we'll make the curve red.

For `k = e^-2` we'll make the curve blue.

So there is the projection of our contours. You could imagine that the z component of each contour increases as we go further up on the y-axis, if you wanted a sense of the curve. Hope that helps!

Sources:

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