Question

In: Math

Function of Several Variables

Compute the directional derivative aong, u u , at the indicated points .

(a). f(x,y)=xy3 f(x,y)=x\sqrt{y-3}  ,u=(1,6) ,u=(-1,6)  a=(2,12) a=(2,12)

(b). f(x,y,z)=1x+2y3z f(x,y,z)=\frac{1}{x+2y-3z} , u=(12,9,4) u=(12,-9,-4) a=(1,1,1) a=(1,1,-1)

Solutions

Expert Solution

solution

Note : The directional derivative along u is :

Duf(x,y)=Fx(x,y)cos(θ)+Fy(x,y)sin(θ) D_uf(x,y)=F_x(x,y)cos(\theta)+F_y(x,y)sin(\theta)

Duf(a)=Fx(a).v1+Fy(x,y).v2 D_uf(a)=F_x(a).v_1+F_y(x,y).v_2

where v(x,y)=uu=v1i+v2j v(x,y)=\frac{u}{||u||}=v_1i+v_2j

(a). f(x,y)=xy3 f(x,y)=x\sqrt{y-3}  ,u=(1,6) ,u=(-1,6)  a=(2,12) a=(2,12)

Let's find the unit vector of u : v=uu=137+637 v=\frac{u}{||u|}=-\frac{1}{\sqrt{37}}+\frac{6}{\sqrt{37}}

Then find Fx(x,y)=y3=>Fx(a)=3 F_x(x,y)=\sqrt{y-3}=>F_x(a)=3

                Fy(x,y)=x2y3=>Fy(a)=13 F_y(x,y)=\frac{x}{2\sqrt{y-3}}=>F_y(a)=\frac{1}{3}

Then, the directional derivative is Duf(a)=337+237=137 D_uf(a)=-\frac{3}{\sqrt{37}}+\frac{2}{\sqrt{37}}=-\frac{1}{\sqrt{37}}

(b)=f(x,y,z)=1x+2y3z f(x,y,z)=\frac{1}{x+2y-3z}  u=(12,9,4) u=(12,-9,-4)   a=(1,1,1) a=(1,1,-1)

Let's find the unit vector of v=uu=1241i+1241j1241 v=\frac{u}{||u|}=\frac{1}{\sqrt{241}}i+\frac{1}{\sqrt{241}} j-\frac{1}{\sqrt{241}}

Let's find the partial derivative of f at ponit a=(1,1,1) a=(1,1,-1)

Fx(x,y,z)=1(x+2y3z)2=>Fx(a)=136 F_x(x,y,z)=-\frac{1}{(x+2y-3z)^2}=>F_x(a)=-\frac{1}{36}

Fy(x,y,z)=2(x+2y3x)2=>Fy(a)=236 F_y(x,y,z)=-\frac{2}{(x+2y-3x)^2}=>F_y(a)=-\frac{2}{36}

Fz(x,y,z)=3(x+2y3z)2=>Fz(a)=336 F_z(x,y,z)=\frac{3}{(x+2y-3z)^2}=>F_z(a)=\frac{3}{36}

Then, the derictional derivative is : Duf(a)=16241 D_uf(a)=-\frac{1}{6\sqrt{241}}


Answe

(a). Therefore, the directional derivative is: Duf(a)=337+237=137 D_uf(a)=-\frac{3}{\sqrt{37}}+\frac{2}{\sqrt{37}}=-\frac{1}{\sqrt{37}}

(b).Therefore, Duf(a)=16241 D_uf(a)=-\frac{1}{6\sqrt{241}}

 

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