Question

Find the directional derivative of the function 5x^2y - 5y^2z + 5/2 (z^2x) at (1,1,1) in the direction of the line a = i + 2j +nk which is parallel to a plane 2(x-5)+3(y-4)+(z-3)=0

Answer 1

Answer:

The unit vector with direction the gradient is:

u=∇f||∇f||.

So you want −u. Comparing two values in your work 815 and 5√8 makes me skeptical.

For the rate of change in direction of −u, we write

D−uf=∇f⋅(−u)=−∇f⋅∇f||∇f||=−||∇f||2||∇f||=−||∇f||.

Answer 2

Answer:

$cos\theta = \frac{20}{\sqrt{8974}}$

Step-by-step explanation:

What is a directional derivative?

• Let $\phi$(x,y,z) be a scalar point function and $\vec{a}$ be a vector inclined at an angle to the direction of ∇$\phi$.
• The component of ∇$\phi$ along $\vec{a}$ is called the directional derivative of $\phi$ along $\vec{a}$.
• This is denoted by   $\frac{\partial{\phi}}{\partial{\vec{a}}}$

Let, $\phi(x,y,z)= 5x^{2}y-5y^{2}z+\frac{5}{2}z^{2}x$

      ∇$\phi=$  $\frac{\partial{\phi}}{\partial{x}} \vec{i}+ \frac{\partial{\phi}}{\partial{y}} \vec{j}+ \frac{\partial{\phi}}{\partial{z}} \vec{k}$   - - - - (1)

     

$\frac{\partial{\phi}}{\partial{x}}= \frac{\partial{(5x^{2}y-5y^{2}z+\frac{5}{2}z^{2}x)}}{\partial{x}} = 10xy+\frac{5}{2}z^{2}$ - - - - - - (2)

$\frac{\partial{\phi}}{\partial{y}}= \frac{\partial{(5x^{2}y-5y^{2}z+\frac{5}{2}z^{2}x)}}{\partial{y}} = 5x^{2}-10yz$ - - - - - - (3)

$\frac{\partial{\phi}}{\partial{z}}= \frac{\partial{(5x^{2}y-5y^{2}z+\frac{5}{2}z^{2}x)}}{\partial{z}} = -5y^{2}+5zx$ - - - - - - (4)

Substitute equations (2),(3) & (4), that is, the derivatives in (1)

       $\bigtriangledown\phi= (10xy+\frac{5}{2}z^{2})\,\vec{i} \,+ \, (5x^{2}-10yz)\,\vec{j}\,+ \, (-5y^{2}+5zx)\,\vec{k}$

$\bigtriangledown\phi_{(1,1,1)} = (10\times1\times1+\frac{5}{2}\times1^{2})\,\vec{i} \,+ \, (5\times1^{2}-10\times1\times1)\,\vec{j}\,+ \, (-5\times1^{2}+5\times1\times1)\,\vec{k}\\\\\\\bigtriangledown\phi_{(1,1,1)} = (10+\frac{5}{2})\,\vec{i} \,+ \, (5-10)\,\vec{j}\,+ \, (-5+5)\,\vec{k}\\\\\\\bigtriangledown\phi_{(1,1,1)} = \frac{25}{2}\,\vec{i} \,+ \, -5\,\vec{j}\,+ \, 0\,\vec{k}$

$|\bigtriangledown\phi|=\sqrt{(\frac{25}{2})^{2} + (-2)^{2}+0^{2}}$  $= \frac{\sqrt{641}}{2}$

Given, $a=\vec{i}+2\vec{j}+n\vec{k}$ is parallel to plane 2(x-5)+3(y-4)+(z-3)=0

Let, $\psi=2(x-5)+3(y-4)+(z-3)$

      $\bigtriangledown\psi=\frac{\partial{\psi}}{\partial{x}} \vec{i}+ \frac{\partial{\psi}}{\partial{y}} \vec{j}+ \frac{\partial{\psi}}{\partial{z}} \vec{k}$

$\frac{\partial{\psi}}{\partial{x}}= \frac{\partial{(2(x-5)+3(y-4)+(z-3))}}{\partial{x}} = 2\\\\\\\frac{\partial{\psi}}{\partial{y}}= \frac{\partial{(2(x-5)+3(y-4)+(z-3))}}{\partial{y}} = 3\\\\\\\frac{\partial{\psi}}{\partial{z}}= \frac{\partial{(2(x-5)+3(y-4)+(z-3))}}{\partial{z}} = 1$

∴  $\bigtriangledown\psi = 2\vec{i}+3\vec{j}+\vec{k}$

$|\bigtriangledown\psi|=\sqrt{(2^{2}+3^{2}+1^{2})}=\sqrt{14}$

Now, $\bigtriangledown\phi \,.\,\bigtriangledown\psi=|\bigtriangledown\phi||\bigtriangledown\psi| \, cos\theta$

 $(\frac{25}{2}\,\vec{i} \,+ \, -5\,\vec{j}\,+ \, 0\,\vec{k})(2\vec{i}+3\vec{j}+\vec{k})=|\,\frac{\sqrt{641}}{2} \,|\,|\,\sqrt{14} \,|\,cos\theta$

                                   $25-15+0=\frac{\sqrt{8974}}{2} cos\theta$

                                        $10 * \frac{2}{\sqrt{8974}}= cos\theta$

                                        $cos\theta = \frac{20}{\sqrt{8974}}$

             

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