Questions with Solutions for Calculus II | MATH 114, Study notes of Mathematics

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Math 114, Chapter 15, Section 6, Question 28

Shuvra Gupta

October 26, 2006

Please refer to the question mentioned above. Let ∇f denote the gradient of f. We agreed ∇f = (2x + y cos(xy))i + (x cos(xy))j. So, in particular, at the point (1, 0), ∇f = 2i + j.

Solution I:

Let u = ai+bj, and since u is a unit vector, we know

√

a^2 + b^2 = 1, which in particular implies a^2 + b^2 = 1. We also know

Duf = ∇f · u = (2i + j) · (ai + bj) = 2 a + b.

But, we are given that Duf = 1, which implies 2a + b = 1. So, b = 1 − 2 a. Using the fact that (a^2 + b^2 ) = 1, and substituting the expression we got for b, we have a^2 + (1 − 2 a)^2 = 1 ⇒ 5 a^2 − 4 a = 0 ⇒ a = 0, (^45) The corresponding values of b are 1 and −^35 respectively. So, the two directions in which the directional derivative equals 1 are j and 45 i − 35 j. I think this method above is the easiest but we can also do it by the other method.

Solution II:

Since u is a unit vector, we can express u as (cos θ)i + (sin θ)j, where θ is the angle u makes with the positive x-axis. Now,

Duf = ∇f · u = (2i + j).(cos θi + sin θj) = 2 cos θ + sin θ.

But, we know Duf = 1, so that implies 2 cos θ + sin θ = 1. So, sin θ = 1 − 2 cos θ. Squaring both sides, we get, sin^2 θ = (1 − 2 cos θ)^2 ⇒ sin^2 θ = 1 + 4 cos^2 θ − 4 cos θ ⇒ 1 − cos^2 θ = 1 + 4 cos^2 θ − 4 cos θ ⇒ 5 cos^2 θ − 4 cos θ = 0. ⇒ cos θ = 0, 45. The corresponding values of sin θ are 1 and −^35 respectively. (Because sin θ = 1 − 2 cos θ.) So, the two directions in which the directional derivative equals 1 are j and 45 i − 35 j.

P.S.: One could also solve this problem using the formula

Duf = ∇f · u = |∇f |(cos α),

where α is the angle between ∇f and u, but this method is messier than the above two. Hope this helps clarify the confusion in class.