# directional derivatives!?

• Nov 6th 2008, 09:42 AM
Ash_underpar
directional derivatives!?
when f(x,y,z)= x^2+y^2-z and r(t)= [t/root2, t/root2, t^2]

find the directional derivative of f along the curve r(t) at the point P where t= root2

i have spent hours trying to work this one out but it is just beyond me! even just to know where to start would be a great help
• Nov 6th 2008, 10:01 AM
Chris L T521
Quote:

Originally Posted by Ash_underpar
when f(x,y,z)= x^2+y^2-z and r(t)= [t/root2, t/root2, t^2]

find the directional derivative of f along the curve r(t) at the point P where t= root2

i have spent hours trying to work this one out but it is just beyond me! even just to know where to start would be a great help

Recall that the directional derivative is defined as $\nabla f(x_0,y_0,z_0)\cdot\bold u$

To find the point $(x_0,y_0,z_0)$ find the component values of $\bold r(t)$ at $t=\sqrt{2}$ Thus, the point is $(1,1,2)$.

Now, lets find $\nabla f(x,y,z)$

$\nabla f(x,y,z)=\left<\frac{\partial}{\partial x}(x^2+y^2-z),\frac{\partial}{\partial y}(x^2+y^2-z), \frac{\partial}{\partial z}(x^2+y^2-z)\right>$ $=\left<2x,2y,-1\right>$

Thus, $\nabla f(1,1,2)=\left<2,2,-1\right>$

Now, let's find the unit vector $\bold u$.

Since $\bold r(\sqrt{2})=\left<1,1,2\right>$, we need to normalize it.

Its magnitude is $\sqrt{1+1+4}=\sqrt{6}$. As a result, $\bold u=\left<\frac{1}{\sqrt{6}},\frac{1}{\sqrt{6}},\fra c{2}{\sqrt{6}}\right>=\left<\frac{\sqrt{6}}{6},\fr ac{\sqrt{6}}{6},\frac{\sqrt{6}}{3}\right>$

Now, find $\nabla f(1,1,2)\cdot \bold u$:

$\nabla f(1,1,2)\cdot \bold u=\left<2,2,-1\right>\cdot\left<\frac{\sqrt{6}}{6},\frac{\sqrt{ 6}}{6},\frac{\sqrt{6}}{3}\right>=\frac{\sqrt{6}}{3 }+\frac{\sqrt{6}}{3}-\frac{\sqrt{6}}{3}=\color{red}\boxed{\frac{\sqrt{6 }}{3}}$

Does this make sense?

--Chris
• Nov 6th 2008, 10:45 AM
Ash_underpar
cheers matey you make it sound so simple but this vector calculus section im doing at the minute is going right over my head!! thanks anyway