# MTH 101 Chapter 7: dis 7 sol

Calculus II Discussion Problems (Week 7) Solutions Spring 2021

1. (a) Answer:

Duf(1,2) = lim

h→0

f1 + 3

5h, 2 + 4

5h−f(1,2)

h= lim

h→0−28

5h−24

25 h2

h=−28

5.

(b) Answer: Since ∇f= (2y)i+ (2x−6y)jand (∇f)(1,2) = 4i−10j, we have

Duf(1,2) = (∇f)(1,2) ·u= (4i−10j)·3

5i+4

5j=12

5−40

5=−28

5.

2. Answer: It can be seen from the graph that fx(0,0) = fy(0,0) = 0. So (∇f)(0,0) = 0. Hence,

for any unit vector u,

(∇f)(0,0) ·u= 0.

i.e., the directional derivative of fat (0,0) is zero in every direction. In particular, it is zero in the

direction of 2i+jas well.

3. (a) Answer: At X0,fincreases most rapidly in the direction of ∇f(X0) = 2i+ 4j+k.

(b) Answer: Since ∇f(X0) = 2i+ 4j+k,fhas a zero rate of change in any direction orthogonal

to 2i+ 4j+k. Using dot product, we can easily ﬁnd vectors vsuch that v·(2i+ 4j+k) = 0.

For instance, the vectors v1= 2i−j,v2=i−2k,v3=j−4kare such vectors. So fhas a zero

rate of change in the directions of v1,v2, and v3. (There are inﬁnitely many other directions

in which fhas a zero rate of change.)

(c) Answer: The vector i+j−kis pointing the same direction as the unit vector u=1

√3i+

1

√3j−1

√3k. Hence

∇fX0·u= (2i+ 4j+k)·1

√3i+1

√3j−1

√3k=5

√3.

(d) Answer: Such a tangent plane is orthogonal to ∇fX0= 2i+ 4j+k. So the tangent plane can

be described by the equation

2(x−1) + 4(y−2) + (z−4) = 0.

4. Answer: The functions fand ghave the same directional derivative in every direction at (a, b).

To see why, note

g(x, y) = f(a, b) + fx(a, b)(x−a) + fy(a, b)(y−b).

So gx(a, b) = fx(a, b) and gy(a, b) = fy(a, b). Thus, (∇f)(a, b) = (∇g)(a, b) and, for any unit vector

u,

Duf(a, b) = ∇f(a, b)·u=∇g(a, b)·u=Dug(a, b).

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