 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2021/paper-1/14/b/solution

The original PDF specifies evaluation at

(1, 1, 1)

; the local TeX incorrectly gives

(1, 1, 0)

. Write the scalar field explicitly as

f = c / (x^{2} + y^{2} + z^{2})

. Its gradient is

\nabla f = - \frac{2 c ( x , y , z )}{( x ^{2} + y ^{2} + z ^{2} ) ^{2}}, \nabla f (1, 1, 1) = - \frac{2 c}{9} (1, 1, 1) .

A directional derivative measures rate per unit distance, so normalize the direction to the unit vector

n = (1, 1, 0) / 2

. Then

D_{n} f (1, 1, 1) = \nabla f (1, 1, 1) \cdot n = - \frac{2 2 c}{9} .

Using the unnormalized direction vector would instead differentiate with respect to a path parameter, not distance, and would give a different scale.

 New to topics? Read the docs here!