MA1312 (Advanced Calculus) Solutions to tutorial sheet 6
[March 5 – 6, 2013]
1. Let f (x, y) = xy sin(2πx + πy) − y 2 .
(a) Find the gradient vector ∇f at a general point (x, y) then compute ∇f |(1/4,−1/2) .
Solution:
∂f
= y sin(2πx + πy) + 2πxy cos(2πx + πy)
∂x
∂f
= x sin(2πx + πy) + πxy cos(2πx + πy) − 2y
∂x
So
∇f = (y sin(2πx+πy)+2πxy cos(2πx+πy), x sin(2πx+πy)+πxy cos(2πx+πy)−2y)
We could use vector notation and write instead
∇f = (y sin(2πx+πy)+2πxy cos(2πx+πy))i+(x sin(2πx+πy)+πxy cos(2πx+πy)−2y)j.
We also have
1
π
1
π
π
π
∇f |(1/4,−1/2) = − sin 0 − cos 0, sin 0 − cos 0 + 1 = − , 1 −
2
4
4
8
4
8
(b) Find the directional derivative Du f (1/4, −1/2) for the function f (x, y), u = (u1 , u2 )
any unit vector.
Solution:
π
π
u2
Du f (1/4, −1/2) = − u1 + 1 −
4
8
(c) Find the direction u in which the directional derivative Du f (1/4, −1/2) is largest
and find this largest, possible value.
Solution:
− π4 , 1 − π8
∇f |(1/4,−1/2)
u =
=
−π, 1 − π k∇f |(1/4,−1/2) k
4
8
!
π
8−π
p
=
, p
2
2
2
4 π /16 + (1 − π/8) 8 π /16 + (1 − π/8)2
!
π
8−π
p
=
, p
π 2 + (1 − π/2)2 2 π 2 + (1 − π/2)2
The largest possible value is
π
π p 2
k∇f |(1/4,−1/2) k = − , 1 −
= π /16 + (1 − π/8)2
4
8
(d) Suppose there is a parametric curve p(t) = (x(t), y(t)) in R2 satisfying p(2) = (1, 1)
and p0 (2) = (1, −1).
d
i. Find
|t=2 f (p(t)).
dt
Solution: Considering (x, y) = p(t) = (x(t), y(t)) the chain rule in 2 variables
says
d
∂f dx ∂f dy
f (x, y) =
+
dt
∂x dt
∂y dt
When we evaluate at t = 2 we have
(x, y) = (x(2), y(2)) = p(2) = (1, 1)
thus
∂f ∂f
,
= ∇f |(1,1) = (sin(3π)+2π cos(3π), sin(3π)+π cos(3π)−2) = (−2π, −π−2)
∂x ∂y
and
dx dy
,
dt dt
= p0 (2) = (1, −1)
So we have
d
|t=2 f (p(t)). = (−2π)(1) − (π + 2)(−1) = −2π + π + 2 = 2 − π
dt
ii. Use the geometric interpretation of the gradient to show that the above parametric representation p(t) could not describe the level curve f (x, y) = −1.
Solution: We proved in class that the gradient is perpendicular to the level
curves, thus, had p(t) described the level curve f (x, y) = −1 we must have
had that ∇f (1, 1) is perpendicular at the point (1,1) to the tangent to the level
curve f (x, y) = −1 (the tangent is the vector p’(2)). However, we compute
∇f (1, 1) · p0 (2) = (−2π, −π − 2) · (1, −1) = 2 − π 6= 0.
Hence, p could not describe the level curve f (x, y) = −1.
2. If g(x, y) = x2 + y 2 − 4x, find the gradient vector ∇g(1, 2) and use it to find the tangent
line to the level curve g(x, y) = 1 at the point (1,2). Sketch the level curve, the tangent
line, and the gradient vector.
Solution: We compute ∇g(1, 2) = (−2, 4) and since the tangent line to the level curve
is perpendicular to the gradient, we construct the tangent at (1,2) with direction (2,-1)
(any vector perpendicular on (-2,4) will do). The level curve g(x, y) = 1 is given by the
equation
x2 − 4x + y 2 = 1
which can be rewritten as (x−2)2 +y 2 √
= 5. We recognize this last equation as the equation
of the circle centered at (2,0) of radius 5, which passes through the given point (1,2). The
gradient will have the direction of the radius to this circle passing through (1,2), with the
tangent to the circle perpendicular to it.
Petronela Radu
2