MA22S1: SOLUTIONS TO TUTORIAL 4
1. Using the Chain Rule find
dw
.
dt
(i) w = x2 + y 2 where x = cos t and y = sin t.
(ii) w = z − sin (xy) where x = t, y = ln t and z = et−1 .
Solution:
(a)
∂w dx ∂w dy
dw
=
+
dt
∂x dt
∂y dt
= 2x(− sin t) + 2y(cos t)
= −2 cos t sin t + 2 cos t sin t
= 0
(b)
dw
∂w dx ∂w dy ∂w dz
=
+
+
dt
∂x dt
∂y dt
∂z dt
1
= −y cos (xy) − x cos (xy) ( ) + et−1
t
= − ln t cos (t ln t) − cos (t ln t) + et−1
2. Using the Chain Rule find
∂w
∂r
and
∂w
.
∂s
(a) w = x2 + xy 2 where x = rs + 1 and y = ers .
(b) w = xy + yz + xz where x = r + s, y = r − s and z = rs
Solution:
1
2
MA22S1: SOLUTIONS TO TUTORIAL 4
(a)
∂w
∂w ∂x ∂w ∂y
=
+
∂r
∂x ∂r
∂y ∂r
= (2x + y 2 )(s) + (2xy)(sers )
= (2(rs + 1) + e2rs )(s) + 2(rs + 1)(se2rs )
∂w
∂w ∂x ∂w ∂y
=
+
∂s
∂x ∂s
∂y ∂s
= (2x + y 2 )(r) + (2xy)(rers )
= (2(rs + 1) + (ers )2 )(r) + 2(rs + 1)(re2rs )
(b)
∂w ∂x ∂w ∂y ∂w ∂z
∂w
=
+
+
∂r
∂x ∂r
∂y ∂r
∂z ∂r
= (y + z) + (x + z) + (x + y)s
= (r − s + rs) + (r + s + rs) + 2rs
= 2r + 4rs
∂w ∂x ∂w ∂y ∂w ∂z
∂w
=
+
+
∂s
∂x ∂s
∂y ∂s
∂z ∂s
= (y + z) − (x + z) + (x + y)r
= (r − s + rs) − (r + s + rs) + 2r 2
= −2s + 2r 2
3. Use implicit differentiation to find
dy
dx
at the point P (0, ln 2) where
xey + sin (xy) + y − ln 2 = 0
MA22S1: SOLUTIONS TO TUTORIAL 4
3
3*cos(u)*cos(v), 3*sin(u)*cos(v), 3*sin(v)
3
2.5
2
1.5
1
0.5
0
-3
-2
-1
0
1
2
-3
3
Figure 1. Graph of f (x, y) =
-2
-1
0
1
2
3
p
9 − x2 − y 2
Solution: First we write F (x, y) = xey + sin (xy) + y − ln 2. We have
Fx (x, y) = ey + y cos(xy)
Fx (0, ln 2) = 2 + ln 2
Fy (x, y) = xey + x cos(xy) + 1
Fy (0, ln 2) = 1
Then
dy
Fx
=−
= −2 − ln 2
dx
Fy
4. The temperature at a point (x, y) is described by the function
f (x, y) =
p
9 − x2 − y 2
(a) Sketch the graph of f and sketch the level curves for k = 0, 1, 2.
(b) What is the rate of change of the temperature at the point
P (2, 1) in a north-easterly direction?
Solution:
(a) See Figure 1 and Figure 2.
4
MA22S1: SOLUTIONS TO TUTORIAL 4
sqrt(8)*cos(t),
sqrt(5)*cos(t),
3*cos(t),
sqrt(8)*sin(t)
sqrt(5)*sin(t)
3*sin(t)
3
2
1
0
-1
-2
-3
-3
-2
-1
0
1
Figure 2. Level curves for f (x, y) =
k = 0, 1, 2
2
3
p
9 − x2 − y 2 and
(b) The rate of change in a north-easterly direction is given by
the directional derivative Du f where we take the unit vector
u = h √12 , √12 i. First we compute the gradient of f (x, y),
∇f (x, y) = hfx (x, y), fy (x, y)i
*
+
x
y
=
−p
, −p
9 − x2 − y 2
9 − x2 − y 2
At P (2, 1) we have
1
∇f (2, 1) = h−1, − i.
2
Thus the directional derivative at P (2, 1) in the direction of
u = h √12 , √12 i is,
Du f (2, 1) = ∇f (2, 1) · u
1 1
1
= h−1, − i · h √ , √ i
2
2 2
1
1
3
= −√ − √ = − √
2 2 2
2 2