MATH 253 WORKSHEET 15
DIRECTIONAL DERIVATIVES
(1) An ant is crawling along the curve
y = x2
vcm/s (distances are measured in cm). The
y
T (x, y) = 1+x
2 . What is the rate of change of
the temperature the ant sees when it is located at (x, y)?
dy
Solution: At (x0 , y0 ) the tangent line to y = x2 has slope 2x0 (the derivative dx
), and so is
parallel to the vectorh1, 2x0 i [for a detailed derivation, the tangent line: has the equation y = y0 +
2x0 (x − x0 ) = 2x0 x−x20 and therefore the parametrization x, 2x0 x − x20 = 0, −x20 +x h1, 2x0 i]. A
unit vector in the direction of travel is therefore ~
u = √ 1 2 h1, 2x0 i. The gradient of the temperature
temperature in the
xy
at the rate of
plane is varying according to
1+4x0
is
~ (x0 , y0 )
∇T
∂T ∂T
,
∂x ∂y
1
−2x0 y0
,
=
(1 + x20 )2 1 + x20
1
=
−2x0 y0 , 1 + x20 .
2
2
(1 + x0 )
=
The directional derivative in the direction
(measuring rate of change of tempreature per unit
~u
distance travelled) is then
1
2
D~u T = p
2 −2x0 y0 , 1 + x0 · h1, 2x0 i ,
2
2
1 + 4x0 (1 + x0 )
and the rate of change (measuring change per unit time) is thus
2x3 + 2x0 − 2x0 y0
vD~u T = p 0
2v.
1 + 4x20 (1 + x20 )
(2) Show that every plane tangent to the surface
Solution:
2
2
z 2 = x2 + y 2
passes through the origin.
2
F (x, y, z) = x + y − z . Then the gradient vector at any point is tangent to
~ (x0 , y0 , z0 ) = h2x0 , 2y0 , −2z0 i, the equation of the plane
the level surface at that point. Since ∇F
tangent at (x0 , y0 , z0 ) where F (x0 , y0 , z0 ) = 0 is therefore
Let
2x0 (x − x0 ) + 2y0 (y − y0 ) − 2z0 (z − z0 ) = 0 ,
or
x0 x + y0 y − z0 z = z02 − x20 − y02 = 0 .
Clearly
Date
(0, 0, 0)
solves this equation.
: 11/10/2013.
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