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Exercise 5

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Exercise 5
BGU Physics Dept. Introduction to Mathematical Methods in Physics
Exercise 5
Multivariable scalar functions
df
1. In the following find
for the composite function f (x(t), y(t), z(t)) in two ways: (i) use the
dt
chain rule, then express your answer in terms of t.
(ii) express the composite function f in terms of t and differentiate.
(a) w = xyz, x = t, y = t2 , z = t3
w = t6 → w = 6t5
(1)
dw
∂w dx ∂w dy ∂w dz
=
+
+
= yz + 2txz + 3t2 xy = t5 + 2t5 + 3t5 = 6t5 (2)
dt
∂x dt
∂y dt
∂z dt
(b) w = ln(x2 + y 2 ), x = 2 cos t, x = 2 sin t
w = ln(4) ⇒ dw/dt = 0
dw
2x
2y
=
(−2 sin t) + 2
(2 cos t) = 0
2
2
dt
x +y
x + y2
(3)
(4)
2. Calculate the mixed second derivatives of the function f (x, y) = xy + 2x − ln(x2 y) and show
∂2f
∂2f
= ∂y∂x
.
that they are equal. That is, show that ∂x∂y
∂2f
2
∂2f
= 2 =
∂x∂y
x
∂y∂x
(5)
3. For the function r(x, y, z) = (x2 + y 2 + z 2 )1/2 , calculate
∂2f ∂2f
,
∂x2 ∂y 2
and
∂2f
∂x∂y .
4. Partial derivatives
y 2 +z 2
(x2 +y 2 +z 2 )3/2
x2 +z 2
∂y2 f = (x2 +y
2 +z 2 )3/2
∂x ∂y f = (x2 +y−xy
2 +z 2 )3/2
(a) ∂x2 f =
5. Stationary points
(a) Find and classify the extremal points of the function f (x, y) = xy in the domain specified
by x ≥ 0, y ≥ 0, 2x + 3y ≤ 10.
(0,0) is the only stationary point, f (0, 0) = 0. Boundaries: If x = 0or y = 0 the function
and f (x, y) = x 10−2x
for 0 ≤ x ≤ 5. The
is zero. If 2x + 3y = 10 then y = 10−2x
3
3
maximum of this parabola is at x = 5/2, hence y = 5/3, and
f (5/2, 5/3) = 25/6 a maximum. All (x, 0)and (0, y)are minima .
1
(b) Find the stationary values of f (x, y) = x2 + xy + y 2 + 3x − 3y + 4 and classify them
as maxima, minima or saddle points. Make a rough sketch of the contours of f in the
quarter plane x, y ≥ 0.
6. (a) y = c/x
~ = (y, x), ∇f
~ (1, 1) = (1, 1)
(b) ∇f
π
4
(anti-parallel to the gradient).
(c) Dv̂ f (1, 1) = cos θ + sin θ, taking a maximum at θ =
a minimum at θ =
5π
4
(parallel to the gradient) and
(d) z − 1 = 1(x − 1) + 1(y − 1) or z = x + y − 1 .
Figure 1: question 3b.
7. Find an equation for the plane that is tangent to the given surface at the given point
z = ln(x2 + y 2 ), (1, 0, 0)⇒ z = 2x − 1, y = 0
z = e
−(x2 +y 2 )
, (0, 0, 1) ⇒ z = 1
(6)
(7)
8. (a) Find the curve of intersection of the surfaces z = x2 −y 2 and z = 2+(x−y)2 in parametric
form.
The curve of intersection is given by y 2 − xy + 1 = 0. Solving for y(x) and setting y = t
we have the parametric form ~r(t) = (x(t), y(t), z(t)) with
x = t+
1
t
(8)
y = t
(9)
1
z = 2+ 2
t
(10)
~r(t) = (t + 1/t, t, 2 + 1/t2 )
(b) Find the angle of intersection of these two surface at the point (2, 1, 3). (The angle of
intersection of the two surfaces is defined to be the angle made by their tangent planes)
2
The angle between the planes is the angle between the planes normal vectors
n̂1 =
(2x, −2y, −1)|(2,1,3) = (4, −2, −1)
(11)
n̂2 =
(2x − 2y, −2x + 2y, −1)|(2,1,3) = (2, −2, −1)
n̂1 · n̂2
−1
cos
≈ 18.980
|n̂1 ||n̂2 |
(12)
∠(n̂1 , n̂2 ) =
(13)
(c) Check that the tangent vector to the curve of intersection found in part (a) at the point
(2, 1, 3) lies in (i.e. is parallel to) the tangent plane of each of the two surfaces. The
tangent vector is
~v = (1 − 1/t2 , 1, −2/t3 )
(14)
At (2, 1, 3) we have y = t = 1 so ~v = (0, 1, −2) which is perpendicular both to n̂1 and n̂2 .
3
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