Linear Programming Lecture Notes - Penn State Personal Web Server

Then the slope of the tangent line is given by:
dy
dx
= −∂z/∂x
∂z/∂y
By zx(x0, y0) we mean ∂z/∂x evaluated at (x0, y0) and by zy(x0, y0) we mean ∂z/∂y evaluated
at (x0, y0). Then the slope of the tangent line to the curve z(x, y) = k at (x0, y0) is:
m = −zx(x0, y0)
zy(x0, y0)
An equation for the tangent line at this point is:
(1.22) y − y0 = m(x − x0)
We can compute a vector that is parallel to this line by taking two points on the line, (x0, y0)
and (x1, y1) and computing the vector (x1 − x0, y1 − y0). We know that:
y1 − y0 = m(x1 − x0)
because any pair (x1, y1) on the tangent line must satisfy Equation 1.22. Thus we have the
vector v = (x1 − x0, m(x1 − x0)) parallel to the tangent line. Now we compute the dot
product of this vector with the gradient of the function:
We obtain:
∇z(x0, y0) = (zx(x0, y0), zy(x0, y0))
∇z(x0, y0) · v = zx(x0, y0) (x1 − x0) + zy(x0, y0) (m(x1 − x0)) =
−zx(x0, y0)
zx(x0, y0) (x1 − x0) + zy(x0, y0)
zy(x0, y0) (x1
− x0) =
zx(x0, y0) (x1 − x0) + (−zx(x0, y0)(x1 − x0)) = 0
Thus, ∇z(x0, y0) is perpendicular to v as we expected from Theorem 1.17
Example 1.20. Let’s demonstrate the previous remark and Theorem 1.17. Consider the
function z(x, y) = x 4 + y 2 + 2xy with a point (x0, y0). Any level curve of the function is
given by: x 4 + y 2 + 2xy = k. Taking the implicit derivative we obtain:
d
dx
x 4 + y 2 + 2xy = k
d
dx
=⇒ 4x 3 + 2y dy
d
+ 2y + 2x
dx dx
Note that to properly differentiate 2xy implicitly, we needed to use the product rule from
calculus. Now, we can solve for the slope of the tangent line to the curve at point (x0, y0)
as:
dy
m = =
dx
−4x30 − 2y0
2y0 + 2x0
Our tangent line is then described the equation:
y − y0 = m(x − x0)
Using the same reasoning we did in the remark, a vector parallel to this line is given by
(x1 − x0, y1 − y0) where (x1, y1) is another point on the tangent line. Then we know that:
y1 − y0 = m(x1 − x0)
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