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!
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`
`y = x^2+lne^-1 = x^2-1`
Again, same as the first case, just shifted down by one.
`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!