Directional derivatives in the direction of the vector
Directional derivatives for two- and three-variable functions
The directional derivative of a multivariable function takes into account the direction (given by the unit vector ???\vec{u}???) as well as the partial derivatives of the function with respect to each of the variables.
In a two variable function, the formula for the directional derivative is
???D_uf(x,y)=a\left(\frac{\partial{f}}{\partial{x}}\right)+b\left(\frac{\partial{f}}{\partial{y}}\right)???
where
???a??? and ???b??? come from the unit vector ???\vec{u}=\langle{a,b\rangle}???
If asked to find the directional derivative in the direction of ???\vec{v}=\langle{c},d\rangle???, we’ll need to convert ???\vec{v}=\langle{c},d\rangle??? to the unit vector using
???\vec{u}=\left\langle\frac{c}{\sqrt{c^2+d^2}},\frac{d}{\sqrt{c^2+d^2}}\right\rangle ???
???\frac{\partial{f}}{\partial{x}}??? is the partial derivative of ???f??? with respect to ???x???
???\frac{\partial{f}}{\partial{y}}??? is the partial derivative of ???f??? with respect to ???y???
In a three variable function, the formula for the directional derivative is
???D_uf(x,y,z)=a\left(\frac{\partial{f}}{\partial{x}}\right)+b\left(\frac{\partial{f}}{\partial{y}}\right)+c\left(\frac{\partial{f}}{\partial{z}}\right)???
where
???a???, ???b??? and ???c??? come from the unit vector ???\vec{u}=\langle{a,b,c\rangle}???
If asked to find the directional derivative in the direction of ???\vec{v}=\langle{d},e,f\rangle???, we’ll need to convert ???\vec{v}=\langle{d},e,f\rangle??? to the unit vector using
???\vec{u}=\left\langle\frac{d}{\sqrt{d^2+e^2+f^2}},\frac{e}{\sqrt{d^2+e^2+f^2}},\frac{f}{\sqrt{d^2+e^2+f^2}}\right\rangle ???
???\frac{\partial{f}}{\partial{x}}??? is the partial derivative of ???f??? with respect to ???x???
???\frac{\partial{f}}{\partial{y}}??? is the partial derivative of ???f??? with respect to ???y???
???\frac{\partial{f}}{\partial{z}}??? is the partial derivative of ???f??? with respect to ???z???
How to find the directional derivatives in the direction of a given vector
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The directional derivative toward a vector at a particular point
Example
Find the directional derivative of ???f(x,y)??? in the direction of ???\vec{v}=\langle1,2\rangle??? at the point ???P(1,-2)???.
???f(x,y)=2x^3+3x^2y+y^2???
We’ll start by converting the given vector to its unit vector form.
???\vec{u}=\left\langle\frac{c}{\sqrt{c^2+d^2}},\frac{d}{\sqrt{c^2+d^2}}\right\rangle ???
???\vec{u}=\left\langle\frac{1}{\sqrt{(1)^2+(2)^2}},\frac{2}{\sqrt{(1)^2+(2)^2}}\right\rangle ???
???\vec{u}=\left\langle\frac{1}{\sqrt{5}},\frac{2}{\sqrt{5}}\right\rangle ???
Now we’ll find the partial derivatives of ???f??? with respect to ???x??? and ???y???.
???\frac{\partial{f}}{\partial{x}}=6x^2+6xy???
and
???\frac{\partial{f}}{\partial{y}}=3x^2+2y???
With the unit vector and the partial derivatives, we have everything we need to plug into our formula for the directional derivative.
???D_uf(x,y)=\frac{1}{\sqrt{5}}\left(6x^2+6xy\right)+\frac{2}{\sqrt{5}}\left(3x^2+2y\right)???
We want to find the directional derivative at the point ???P(1,-2)???, so we’ll plug this into the equation we just found for the directional derivative, and we’ll get
???D_uf(1,-2)=\frac{1}{\sqrt{5}}\left[6(1)^2+6(1)(-2)\right]+\frac{2}{\sqrt{5}}\left[3(1)^2+2(-2)\right]???
???D_uf(1,-2)=\frac{-6}{\sqrt{5}}+\frac{-2}{\sqrt{5}}???
???D_uf(1,-2)=\frac{-8}{\sqrt{5}}???
This is the directional derivative of the function ???f(x,y)=2x^3+3x^2y+y^2??? in the direction ???\vec{v}=\langle1,2\rangle??? at the point ???P(1,-2)???.