Mathematics 126 test 4 spring 2001 solutions
(1) Given the function f(x,y) = ( x2 – 3y )2,
(a) sketch level curves corresponding to function values f = 0 , f = 1 , and f = 4.
f(x,y) = C
( x2 – 3y )2 = C
x2 – 3y = ±
C
2
y =
x ± C
3
level curves of (x^2-3y)^2
2
f=4
1.5
f=1
1
f=0
0.5
f=1
0
f=4
-0.5
-1
-2
-1.5
-1
-0.5
0
x values
0.5
∂f
∂f
(b) Calculate the first partial derivatives ∂x and ∂y .
∂f
2
2
∂x = 2( x – 3y )(2x) = 4x( x – 3y )
∂f
2
2
∂y = 2( x – 3y )(-3) = -6( x – 3y )
1
1.5
2
(c) Find the gradient vector f at the point P(2,1).
∂f
f = < ∂x
∂f
∂y > (2,1) = < 8 , -6 >
(d) Explain the signifigance of the direction and magnitude of the gradient
vector at the point P(2,1).
The gradient points in the direction in which the values of the function increase
most rapidly (i.e. the direction of the largest directional derivative)
The magnitude of the gradient is the value of the largest directional derivative.
(e) Determine an equation for the tangent plane through the
point (2,1,f(2,1)) = (2,1,1).
∂f
∂f
z = f(2,1) + ∂x
(x - 2) + ∂y
(y - 1)
(2,1)
(2,1)
z = 1 + 8(x - 2) - 6(y - 1)
(f) Find the differential df at the point P(2,1).
∂f
∂f
df = ∂x
dx + ∂y
dy
(2,1)
(2,1)
df = 8dx - 6dy
(g) Use part (e) or (f) to estimate the value of f(1.98,1.03).
part (e) calculation:
f(1.98,1.03)  z
tangent plane
= 1 + 8(1.98 - 2) - 6(1.03 - 1)
= 1 + 8(-0.02) - 6(0.03) = 0.66
part (f) calculation:
f(1.98,1.03)  f(2,1) + df = 1 + 8(dx) - 6(dy)
= 1 + 8(-0.02) - 6(0.03) = 0.66
(h) Calculate the directional derivative of f at the point P(2,1) in the direction of
the vector w = < 3, –4 >.
Dwf(2,1) = compwf =
f . w
w
=
< 8 -6 > . < 3 -4 >
< 3 -4 >
(i) Suppose that the coordinates x and y are functions of two
∂f
variables r and . Write out the chain rule formula for ∂r ,
∂f
and calculate the value of ∂r at the point P(2,1).
48
= 5 = 9.6
∂f
∂f ∂x
∂f ∂y
=
+
∂r
∂x ∂r
∂y ∂r
(note: x = rcos and y = rsin )
∂f
∂f
∂f
=
(cos)
+
∂r
∂x
∂y (sin)
( for the point x = 2 and y = 1, the polar coordinate r =
x
2
y
1
therefore cos = r =
& sin = r =
)
5
5
∂f
10
2
1
=
8
6
=
= 2 5




∂r
5
 5
 5
5.
(j) If a particle moved through the point P(2,1) with velocity v = < 3 , –1 >, at
what rate would the function values of f be changing with respect to time t ?
df
∂f dx
∂f dy
=
+
dt
∂x dt
∂y dt = 8(3) - 6(-1) = 30
(k) Suppose that x and y are linear functions of time:
x = 2 + at and y = 1 + bt , where < a , b > is a unit vector
Use the chain rule (twice) to find a formula for the second derivative, f(t) in
terms of fxx, fxy, fyy, a and b.
First calculate the first derivative:
df
∂f dx
∂f dy
=
+
dt
∂x dt
∂y dt = fx a + fy b
then the second derivative:
d df
∂ df
∂ df
f(t) = dt dt = ∂x dt a + ∂y dt b
 
 
 
∂
∂
= ∂x fx a + fy b (a) + ∂y fx a + fy b (b)
= ( fxx a + fyx b )(a) + ( fxy a + fyy b )(b)
(
)
(
)
= a2 fxx + ab fyx + ab fxy + b2 fyy
(2) Find an equation for the tangent plane to the surface
3
2 4
5 3
x yz – y z + x y = 1
at the point (x,y,z) = (1,1,1).
this is a level surface for the function F(x,y,z) = x3yz – y2z4 + x5y3
F(x,y,z) = x3yz – y2z4 + x5y3 = 1
the gradient F = < Fx , Fy , Fz > is perpendicular to the level surface (and to the
tangent plane to the surface)
F = < 3x2yz + 5x4y3
x3z - 2yz4 + 3x5y2
x3y - 4y2z3 >
(1,1,1)
= < 8 , 3 , -3 >
the tangent plane equation is therefore
8(x-1) + 3(y-1) - 3(z-1) = 0
(3) Using the following contour map below,
(a) sketch the gradient vector at the points P, Q, and R.
Each gradient vector is perpendicular to the level curve at the point.
f
The gradient at P has a magnitude approximately equal to
in a direction
s
f
4-3
perpendicular to the level curve: f 
= .25 = 4
s
Since Q is local maximum point, f = < fx , fy > = < 0 , 0 > = 0
The gradient at Q has a magnitude approximately equal to 
f
1-0
= (1/3) = 3
s
(b) compute the approximate directional derivative in the direction of
the vector < 1 , 1 > at the point P.
Moving in the direction of < 1 , 1 > from the point P, the function values increase
from 3 to 4 in a distance of approximately one unit. Therefore the directional
f
derivative, approximately equal to
, is close to 1.
s
(c) compute an approximate equation for the tangent plane through the
point (2,3,f(2,3)).
At the point (2,3) the gradient points in a direction close to < -1 , 1 >/ 2 , and the
magnitude is approximately 4 (from part a). Hence at (2,3)
fx = -4/ 2 and fy = 4/ 2
Also f(2,3) = 3. The tangent plane equation is approximately:
z = 3 - (4/ 2 )(x-2) + (4/ 2 )(y-3)