Question

By about how much does the function f(x, y, z) =e^x cos(y-z) change as a point P(x, y, z) moves away from the origin a distance of ds= 0.1 in the direction of i+ 2j+3k? (Hint: Consider how we can get the differential df from the differential ds.)

Answer 1

Let us determine the coordinates of the unit vector that is parallel to $\lbrace1,2,3 \rbrace$ :

$\vec{s} = \lbrace c, 2c, 3c\rbrace.$ The modulus is 1, so $\sqrt{c^2+(2c)^2+(3c)^2} = 1 \Rightarrow c = \dfrac{1}{\sqrt{14}}$ .

Therefore, the directional derivative in (0,0,0) will be

${\displaystyle \nabla _{\mathbf {e} }{f}(\mathbf {x} )={\frac {\partial f}{\partial x}}e_{1}+\frac {\partial f}{\partial y}e_{2} +{\frac {\partial f}{\partial z}}e_{3}} = e^x\cos(y-z)\cdot c - e^x\sin(y-z)\cdot 2c + e^x\sin(y-z)\cdot 3c= e^x\cdot c\cdot\big(\cos(y-z) - 2\sin(y-z)+3\sin(y-z)\big) = e^x\cdot c\cdot(\cos(y-z)+\sin(y-z)) = e^0\cdot \dfrac{1}{\sqrt{14}} \cdot\big( \cos 0 + \sin 0\big) = \dfrac{1}{\sqrt{14}}$Next, we should calculate $df = ds\cdot \nabla _{\mathbf {e} }{f}(\mathbf {x} ) = 0.1\cdot \dfrac{1}{\sqrt{14}} .$