Homework week 6
1. Find the gradient, ∇ f , of the function f at the given point. Then sketch the level curves that passes
through the point
a) f x y y x, 2 1 . Solution:
the gradient is
i ∇f
and
∇f 21
j
i j
the level curves are given by y x c which are staright line of slope m 1 (45 ) and that cross
the y axis at c. See fig1.
y
y=2
f
(2,1)
∆
y=1
x
c=0
x=1
x=2
Figure 0.1: fig1. Notice the gradient at 2 0 goes -1 in the x direction and +1 in the y direction.
b) f x y x2 y2 ,
The gradient is
2
2
2 1 . Solution:
∇f
and
∇ f 21
xi yj
2i j
the level curves are c x2 y2 which is the equation of the hyperbola. For c 0 we obtain two
straight lines y x crossing at the origin, the asymptotes (slope m 1 implies 45 , m 1
implies 135 ). The other curve follow the the asymptotes we skech them in fig2.
2
2
2. Find the derivative of the function at P in the direction of u
1
y
f
∆
( 2 ,1)
1
1
2
x
3
Figure 0.2: fig2
a) f x y 2xy 3y2 , P The direction of u is
5 5 , u 4i 3j. Solution:
v
u
u
∇f
The gradient of the function is
4i 3j
42 32
2yi 4
3
i
j
5
5
2x 6y j
evaluated at the point
∇f 2 5 i 55
the directional derivative is then
10i 20j
∇f v b) f x y 2x2 y2 , P c) f
1 1 , u 3i x y xy yz zx, P 1 1 2 , u 2 5
6 5 j
40
60
i
j
5
5
4j
3i 6j 2k
3. Find the directions in which the function increase or decrease most rapidly at the point P. 1
1 Note:
i)The function increases most rapidly if ∇ f vincrease
to another if it is proportional to it so
∇ f that means that v
λ∇ f
v
increase is parallel
to the gradient. A vector is parallel
increase
the direction of vincrease must be also unitary by definition. To make vincrease unitary we chose λ
vincrease
ii)The function decreases most rapidly if ∇ f vdecrease
∇f ∇f
so
∇ f . Following the previous argument
∇ f v
∇f
increase
2
1
∇f
a) f x y x2 xy y2 , P 1 1 . Solution:
To solve the problem we calculate:
∇f
2x y i x 2y j
therefore
∇f 11
2 1 i i
j
i j
1 2 12
1 2 j
the direction of maximum increase is
vincrease
vdecrease
1
i
2
1
i
2
1
j
2
1
j
2
b) f x y x2 y exy sin y, P 1 0 . Solution:
The gradient is:
∇ f 2xy yexy i 2x2 xexy exy cos y j
and so
∇f 10
2 0 i 2 1 1 j
2i 4j
the direction in which f increases most rapidely is given by
vincrease
2i 4j
22 42
2
i
20
2
i
20
1
i j
y
4
j
20
de direction of maximum decrease is
vincrease
c) f x y z xy
Same way:
yz, P 4
j
20
4 1 1 . Solution:
∇f
4. find the tangent plane and the normal line at the point P of the given surface.
a) x2 y2 z2 3, P 1 1 1 .Solution:
The surface is the graph of the function z 3 x2 y2 (this is obtained by factorising z from
the problem’s function), and the equation of the plane tangent to the surface f at the point P, we
know is given by
z z0 fx P ∆x fy P ∆y
where ∆x x0 x and ∆y y y0 . We calculate the derivatives at the point x 0 y0 z0 z0
fx fy 1
111
111
3
x
3
x2
y2
1
1
111
y
3 x 2 y2 111
1 1 1
therefore the equation of the plane tangent to the function at the point 1 1 1 is
1
z
1
1
x x0 1 x 3 x y
1
1 y y0 1 y 1
b) x2 y2 z2 18, P 3 5 4 . Solution:
We can solve this type of problems in two ways, we use the second here. Define the surface level
F x y z x2 y2 z2 18 0
The tangent plane to the surface is
Fx P ∆x Fy P ∆y Fz P ∆z 0
the gradients are
Fx Fy Fz therefore the tangent plane at 3 5 35
4
35
4
35
4
4
6 x 6
2x 2y 2z 35
4
35
4
35
4
6
10
8
has the equation
10 y 5 8 z 4
0
or
6x 36 10y 50
8
3
5
59
x
y
4
4
4
z
4
c) x y z 1, P 0 1 0 . Solution:
We compare the two methods here (you only needed to use one method thought) : the functions
are
z f x y 1 y x
and
F x y z x y z 1 0
The gradients:
fx fy 0
z0
010
010
1
1
010
010
010
and
Fx Fy Fz 1
010
010
010
4
1
1
010
010
1
1
1
1
1
respectively. The equations of the planes are
z0 z
fx P ∆x 0 1 x 0
x y 1
fy P ∆y
1 y 1
and
Fx P ∆x Fy P ∆y Fz P ∆z
0
z
1 x 0
x y 1
5
1 y 1
1 z 0