To find $a,b,c$ as the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , is atmost $64$...
How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $nabla f(1,2,-1) . (1,2,alpha)=64sqrt{1^2+2^2+alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance .
analysis multivariable-calculus partial-derivative
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How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $nabla f(1,2,-1) . (1,2,alpha)=64sqrt{1^2+2^2+alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance .
analysis multivariable-calculus partial-derivative
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34
add a comment |
How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $nabla f(1,2,-1) . (1,2,alpha)=64sqrt{1^2+2^2+alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance .
analysis multivariable-calculus partial-derivative
How to find $a,b,c$ such that the directional derivative of $f(x,y,z)=axy^2+byz+cz^2x^3$ , at $(1,2-1)$ , has a maximum value of $64$ in a direction parallel to $z$-axis ? I think I have to equate $nabla f(1,2,-1) . (1,2,alpha)=64sqrt{1^2+2^2+alpha^2}$ , but I am not sure and I don't even know how to proceed from here even if it is correct . Please help . Thanks in advance .
analysis multivariable-calculus partial-derivative
analysis multivariable-calculus partial-derivative
edited Oct 28 '15 at 4:04
asked Oct 28 '15 at 4:02
user228168
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 2 days ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34
add a comment |
A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34
A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34
add a comment |
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The directional derivative in the z-direction is just $partial f/partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
add a comment |
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The directional derivative in the z-direction is just $partial f/partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
add a comment |
The directional derivative in the z-direction is just $partial f/partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
add a comment |
The directional derivative in the z-direction is just $partial f/partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
The directional derivative in the z-direction is just $partial f/partial z$ (or in the opposite direction, which would just be the negative of that). So you just need to compute that, evaluate it at the desired point, and find the conditions on the constants which ensure it is less than 64.
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
answered Oct 28 '15 at 4:13
Matt Dickau
1,399313
1,399313
add a comment |
add a comment |
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A direction (unit vector) parallel to the z-axis is (0,0,1), not (1,2,$alpha$). Be careful not to confuse directions and position vectors.
– Matt Dickau
Oct 28 '15 at 4:08
@MattDickau : Parallel to the $z$-axis , not exactly on $z$-axis ..
– user228168
Oct 28 '15 at 4:38
$mathbf r(alpha) = (1,2,alpha)$ parameterizes the position vectors of a line parallel to the $z$-axis, but the direction parallel to the $z$-axis is a vector which is pointed along this line. That is, the head and tail are both on the line, so the direction $mathbf d = (1,2,alpha_2) - (1,2,alpha_1) = (0,0,alpha_2 - alpha_1)$. Then, since it is the direction and not magnitude that we care about, we just normalize it to get $(0,0,1)$ or $(0,0,-1)$ depending on whether it points up or down.
– Matt Dickau
Oct 28 '15 at 15:34