MATH 251 – LECTURE 9
JENS FORSGÅRD
http://www.math.tamu.edu/~jensf/
This week: 12.4–6
webAssign: 12.4–6, due 2/15 11:55 p.m.
Friday:
Kevin.
Next week: 12.7
webAssign: 12.7, opens 2/15 12 a.m.
Friday 2/19:
Covering chapters 11 and 12.
Help Sessions:
M W 5.30–8 p.m. in BLOC 161
Office Hours:
BLOC 641C
M 12:30–2:30 p.m.
W 2–3 p.m.
or by appointment.
Tangent planes
Consider the surface defined by z = f (x, y). Let P = P (x0, y0, z0) be a point on this surface. We want to find
an equation for the tangent plane of the surface at the point P . We have the two space curves intersecting at
the point P .
r1(x) = hx, y0, f (x, y0)i and r2(y) = hx0, y, f (x0, y)i
They have tangent vectors at P given by
r01(x0) = h1, 0, fx0 (x0, y0)i and r02(y0) = h0, 1, fy0 (x0, y0)i.
Tangent planes
Therefor, a normal vector to the tangent space is given by
r01(x0) × r02(y0) = h1, 0, fx0 (x0, y0)i × h0, 1, fy0 (x0, y0)i =
Tangent planes
We can conclude that the equation of the tangent plane is of the form
(x − x0)fx0 (x0, y0) + (y − y0)fy0 (x0, y0) − (z − z0) = 0
where
z0 = f (x0, y0).
Exercise 1. Find the equation of the tangent plane to the surface f (x, y) = x2 + ey + 1 at the point (1, 0, 3).
Increments and differentials
Let z = f (x, y). If x and y are given increments ∆x and ∆y, then the increment of z is
∆z = f (x + ∆x, y + ∆y) − f (x, y).
Let z = f (x, y). The differential dz is defined as dz = fx0 (x, y)dx + fy0 (x, y)dy.
Set dx = ∆x and dy = ∆y. If these are small and f is differentiable, then dz ≈ ∆z.
Increments and differentials
Exercise 2. Let z = f (x, y) = x2 + 2y. Compute the increment of f at the point (1, 2) if x and y are given
increments 1/10 and 1/100.
Exercise 3. Let z = f (x, y) = x2 + 2y. Compute the differential of f at the point (1, 2).
Increments and differentials
The approximation dz ≈ ∆z = f (x + ∆x, y + ∆y) − f (x, y), can be rewritten as
f (x + ∆x, y + ∆y) ≈ f (x, y) + dz.
Exercise 4. Give a rational approimation to the number
q
120
168
which is better than
q
121
169
=
11
13 .
Increments and differentials
Exercise
√
√5. Give a rational approximation of the number
169 + 100 = 11 − 13 + 10 = 8.
√
120 −
√
168 +
√
98 which is better than
√
121 −