Find the directional derivative of h(x,y,z) = cos(xy) + e^(yz) + ln(zx) at P0(1, 0, 1/2) in the direction of u = <1, 2, 2>.
|Mathematics
Find the directional derivative of h(x,y,z) = cos(xy) + e^(yz) + ln(zx) at P0(1, 0, 1/2) in the direction of u = <1, 2, 2>.
ScanSolve AI Answer
Answer
2
To find the directional derivative of h(x,y,z) at P0 in the direction of u, we use the formula Duh(P0)=∇h(P0)⋅u^, where u^ is the unit vector in the direction of u.
Step 1: Find the partial derivatives of h(x,y,z) with respect to x, y, and z.
Given h(x,y,z)=cos(xy)+eyz+ln(zx).
Step 2: Form the gradient vector ∇h(x,y,z).
∇h(x,y,z)=⟨−ysin(xy)+x1,−xsin(xy)+zeyz,yeyz+z1⟩
Step 3: Evaluate the gradient at the given point P0(1,0,1/2).
Substitute x=1, y=0, z=1/2 into the gradient vector:
∂x∂h(1,0,1/2)=−(0)sin(1⋅0)+11=0+1=1∂y∂h(1,0,1/2)=−(1)sin(1⋅0)+(21)e(0)(1/2)=0+21e0=21(1)=21∂z∂h(1,0,1/2)=(0)e(0)(1/2)+1/21=0+2=2
So, the gradient at P0 is:
∇h(1,0,1/2)=⟨1,21,2⟩
Step 4: Find the unit vector u^ in the direction of u=i+2j+2k.
First, find the magnitude of u:
∣∣u∣∣=12+22+22=1+4+4=9=3
Now, find the unit vector:
u^=∣∣u∣∣u=31(i+2j+2k)=⟨31,32,32⟩
Step 5: Calculate the directional derivative by taking the dot product of ∇h(P0) and u^.
Du^h(P0)=∇h(1,0,1/2)⋅u^Du^h(P0)=⟨1,21,2⟩⋅⟨31,32,32⟩Du^h(P0)=(1)(31)+(21)(32)+(2)(32)Du^h(P0)=31+62+34Du^h(P0)=31+31+34Du^h(P0)=31+1+4=36=2
The directional derivative is 2.
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To find the directional derivative of h(x,y,z) at P_0 in the direction of u, we use the formula D_uh(P_0) = h(P_0) · u, where u is the unit vector in the direction of u.
Find the directional derivative of h(x,y,z) = cos(xy) + e^(yz) + ln(zx) at P0(1, 0, 1/2) in the direction of u = <1, 2, 2>.
This mathematics problem involves applying core mathematical principles and formulas. Below you will find a complete step-by-step solution with detailed explanations for each step, helping you understand not just the answer but the method behind it.
ScanSolve AI Answer
To find the directional derivative of h(x,y,z) at P_0 in the direction of u, we use the formula D_uh(P_0) = h(P_0) · u, where u is the unit vector in the direction of u. Step 1: Find the partial derivatives of h(x,y,z) with respect to x, y, and z. Given h(x,y,z) = (xy) + e^yz + (zx). ( h)/( x) = ()/( x)((xy)) + ()/( x)(e^yz) + ()/( x)((zx)) ( h)/( x) = -y(xy) + 0 + (1)/(zx) · z = -y(xy) + (1)/(x) ( h)/( y) = ()/( y)((xy)) + ()/( y)(e^yz) + ()/( y)((zx)) ( h)/( y) = -x(xy) + e^yz · z + 0 = -x(xy) + ze^yz ( h)/( z) = ()/( z)((xy)) + ()/( z)(e^yz) + ()/( z)((zx)) ( h)/( z) = 0 + e^yz · y + (1)/(zx) · x = ye^yz + (1)/(z) Step 2: Form the gradient vector h(x,y,z). h(x,y,z) = -y(xy) + (1)/(x), -x(xy) + ze^yz, ye^yz + (1)/(z) Step 3: Evaluate the gradient at the given point P_0(1, 0, 1/2). Substitute x=1, y=0, z=1/2 into the gradient vector: ( h)/( x)(1,0,1/2) = -(0)(1 · 0) + (1)/(1) = 0 + 1 = 1 ( h)/( y)(1,0,1/2) = -(1)(1 · 0) + ((1)/(2))e^(0)(1/2) = 0 + (1)/(2)e^0 = (1)/(2)(1) = (1)/(2) ( h)/( z)(1,0,1/2) = (0)e^(0)(1/2) + (1)/(1/2) = 0 + 2 = 2 So, the gradient at P_0 is: h(1, 0, 1/2) = 1, (1)/(2), 2 Step 4: Find the unit vector u in the direction of u = i + 2j + 2k. First, find the magnitude of u: ||u|| = sqrt(1^2 + 2^2 + 2^2) = sqrt(1 + 4 + 4) = sqrt(9) = 3 Now, find the unit vector: u = u||u|| = (1)/(3)(i + 2j + 2k) = (1)/(3), (2)/(3), (2)/(3) Step 5: Calculate the directional derivative by taking the dot product of h(P_0) and u. D_uh(P_0) = h(1, 0, 1/2) · u D_uh(P_0) = 1, (1)/(2), 2 · (1)/(3), (2)/(3), (2)/(3) D_uh(P_0) = (1)((1)/(3)) + ((1)/(2))((2)/(3)) + (2)((2)/(3)) D_uh(P_0) = (1)/(3) + (2)/(6) + (4)/(3) D_uh(P_0) = (1)/(3) + (1)/(3) + (4)/(3) D_uh(P_0) = (1+1+4)/(3) = (6)/(3) = 2 The directional derivative is 2. That's 2 down. 3 left today — send the next one.
