Question

Find the directional derivative of x^2,y^2,z^2 at the point (1,1,-1) in the direction of the tangent to the curve r=(e^t)i+ (sin t +1)j + (1 - cos t) at t=0

Answer 1

Answer:

Find the directional derivative of x^2,y^2,z^2 at the point (1,1,-1) in the direction of the tangent to the curve r=(e^t)i+ (sin t +1)j + (1 - cos t) at t=0

Answer 2

The answer is $\bf 2 \sqrt 2$.

Given:

$r=(e^t)i+ (\sin t +1)j + (1 - \cos t)k$ at t = 0.

To Find:

The directional derivative of $x^2+y^2+z^2$ at the point (1,1,-1) along the tangent curve

Solution:

Therefore the tangent of the curve is given by

$t= \frac{\vec r}{|\vec r|}$

at t= 0,

$r =1i+1j+0k\\|r| = \sqrt{1^2+1^2}=\sqrt2$

Therefore the tangent of the curve is

$t=\frac{1}{\sqrt 2}i+\frac{1}{\sqrt{2} }j$

The directional directive of the function will be given by

$\nabla f .t$

$\nabla f = \frac{df}{dx}i +\frac{df}{dy}j +\frac{df}{dz}k\\ \\\nabla f = 2xi+2yj+2zk\\\\\nabla f.t=[2xi+2yj+2zk].[\frac{1}{\sqrt 2}i+\frac{1}{\sqrt{2} }j]\\\\\nabla f.t =\sqrt{2}x+\sqrt 2 y$

Hence for x = 1, y = 1,

$\nabla f.t = 2\sqrt 2$

The directional derivative of the function along the curve is $\bf 2 \sqrt 2$.

#SPJ3