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Calculus II: 3d graph question
The role of sequences in calculusDifferential calculus - Reviewing and drawing graphQuestion regarding calculus, graph of functions, point of inflection.Review of calculus course over the breakcross sections/volume calculusCan't understand logic behind review answer.Differences between directional derivativesApproximate using differentials when partial derivatives are given?Tangent of a Straight LineHow to prove that $fraca-bsqrt1+a^2sqrt1+b^2 < arctana- arctanb$ when $0<b<a$?
$begingroup$
The graph of $z = f (x, y)$ is shown below. In each part, determine whether the given partial derivatives are positive, negative, or zero. (Note that the function is symmetric about 0 in both the x- and y- directions.)
Graph for Problem #2
(a) $f_x(2, −2)$ and $f_xx(2, −2)$
(b) $f_y(2, −2)$ and $f_yy(2, −2)$
(c) $f_x(−2, 0)$ and $f_xx(−2, 0)$
(d) $f_y(−2, 0)$ and $f_yy(−2, 0)$
I have no idea how to solve this question. I am so sorry to just ask like this but our math teacher just went over this topic very briefly and I couldn't find an example question similar to this.
calculus
edited Mar 30 at 3:32
user1952500
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
$endgroup$
add a comment |
$begingroup$
The graph of $z = f (x, y)$ is shown below. In each part, determine whether the given partial derivatives are positive, negative, or zero. (Note that the function is symmetric about 0 in both the x- and y- directions.)
Graph for Problem #2
(a) $f_x(2, −2)$ and $f_xx(2, −2)$
(b) $f_y(2, −2)$ and $f_yy(2, −2)$
(c) $f_x(−2, 0)$ and $f_xx(−2, 0)$
(d) $f_y(−2, 0)$ and $f_yy(−2, 0)$
I have no idea how to solve this question. I am so sorry to just ask like this but our math teacher just went over this topic very briefly and I couldn't find an example question similar to this.
calculus
edited Mar 30 at 3:32
user1952500
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
$endgroup$
$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
2
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42
add a comment |
$begingroup$
The graph of $z = f (x, y)$ is shown below. In each part, determine whether the given partial derivatives are positive, negative, or zero. (Note that the function is symmetric about 0 in both the x- and y- directions.)
Graph for Problem #2
(a) $f_x(2, −2)$ and $f_xx(2, −2)$
(b) $f_y(2, −2)$ and $f_yy(2, −2)$
(c) $f_x(−2, 0)$ and $f_xx(−2, 0)$
(d) $f_y(−2, 0)$ and $f_yy(−2, 0)$
I have no idea how to solve this question. I am so sorry to just ask like this but our math teacher just went over this topic very briefly and I couldn't find an example question similar to this.
calculus
edited Mar 30 at 3:32
user1952500
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
$endgroup$
The graph of $z = f (x, y)$ is shown below. In each part, determine whether the given partial derivatives are positive, negative, or zero. (Note that the function is symmetric about 0 in both the x- and y- directions.)
Graph for Problem #2
(a) $f_x(2, −2)$ and $f_xx(2, −2)$
(b) $f_y(2, −2)$ and $f_yy(2, −2)$
(c) $f_x(−2, 0)$ and $f_xx(−2, 0)$
(d) $f_y(−2, 0)$ and $f_yy(−2, 0)$
I have no idea how to solve this question. I am so sorry to just ask like this but our math teacher just went over this topic very briefly and I couldn't find an example question similar to this.
calculus
calculus
edited Mar 30 at 3:32
user1952500
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
edited Mar 30 at 3:32
user1952500
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
edited Mar 30 at 3:32
user1952500
915712
edited Mar 30 at 3:32
user1952500
915712
edited Mar 30 at 3:32
user1952500
915712
915712
asked Mar 30 at 3:26
curiousengcuriouseng
315
asked Mar 30 at 3:26
curiousengcuriouseng
315
asked Mar 30 at 3:26
curiousengcuriouseng
315
315
$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
2
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42
add a comment |
$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
2
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42
$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
2
2
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints:
- The graph looks like $z=y^2-x^2$.
- For example for $f_colorgreenx(colorgreen2,colorblue-2)$ you imagine a plane parallel to the $colorgreenxz$-plane which is positioned at $colorbluey=-2$.
- This plane intersects with the graph and cuts off the curve $z = colorblue(-2)^2-x^2 = 4-x^2$.
- Now, check slope and convexity of the graph of $z= 4-x^2$ at $colorgreenx = 2$.
- Be careful while using your picture as the $x$-axis is reversely scaled.
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
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1 Answer
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votes
1 Answer
1
active
oldest
votes
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oldest
votes
$begingroup$
Hints:
- The graph looks like $z=y^2-x^2$.
- For example for $f_colorgreenx(colorgreen2,colorblue-2)$ you imagine a plane parallel to the $colorgreenxz$-plane which is positioned at $colorbluey=-2$.
- This plane intersects with the graph and cuts off the curve $z = colorblue(-2)^2-x^2 = 4-x^2$.
- Now, check slope and convexity of the graph of $z= 4-x^2$ at $colorgreenx = 2$.
- Be careful while using your picture as the $x$-axis is reversely scaled.
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
Hints:
- The graph looks like $z=y^2-x^2$.
- For example for $f_colorgreenx(colorgreen2,colorblue-2)$ you imagine a plane parallel to the $colorgreenxz$-plane which is positioned at $colorbluey=-2$.
- This plane intersects with the graph and cuts off the curve $z = colorblue(-2)^2-x^2 = 4-x^2$.
- Now, check slope and convexity of the graph of $z= 4-x^2$ at $colorgreenx = 2$.
- Be careful while using your picture as the $x$-axis is reversely scaled.
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
$endgroup$
add a comment |
$begingroup$
Hints:
- The graph looks like $z=y^2-x^2$.
- For example for $f_colorgreenx(colorgreen2,colorblue-2)$ you imagine a plane parallel to the $colorgreenxz$-plane which is positioned at $colorbluey=-2$.
- This plane intersects with the graph and cuts off the curve $z = colorblue(-2)^2-x^2 = 4-x^2$.
- Now, check slope and convexity of the graph of $z= 4-x^2$ at $colorgreenx = 2$.
- Be careful while using your picture as the $x$-axis is reversely scaled.
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
$endgroup$
Hints:
- The graph looks like $z=y^2-x^2$.
- For example for $f_colorgreenx(colorgreen2,colorblue-2)$ you imagine a plane parallel to the $colorgreenxz$-plane which is positioned at $colorbluey=-2$.
- This plane intersects with the graph and cuts off the curve $z = colorblue(-2)^2-x^2 = 4-x^2$.
- Now, check slope and convexity of the graph of $z= 4-x^2$ at $colorgreenx = 2$.
- Be careful while using your picture as the $x$-axis is reversely scaled.
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
edited Mar 30 at 4:48
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
answered Mar 30 at 4:34
trancelocationtrancelocation
13.6k1829
13.6k1829
add a comment |
add a comment |
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$begingroup$
You need equation of curve. We can't find from graph!
$endgroup$
– Tojrah
Mar 30 at 3:29
$begingroup$
This is all the question gives. I was expecting an equation too ??
$endgroup$
– curiouseng
Mar 30 at 3:31
$begingroup$
This might help youtube.com/…
$endgroup$
– user10478
Mar 30 at 3:33
2
$begingroup$
You don't need the equation for the surface. For example, find the point on the surface corresponding to $(2,-2)$. Is the slope along the $x$ direction positive, negative or zero? What about the convexity?
$endgroup$
– John Douma
Mar 30 at 3:34
$begingroup$
Can you please explain it a bit further? For example there are two different (2,-2) for partial derivative respect to x and partial derivative respect to y. How can I evaluate these two?
$endgroup$
– curiouseng
Mar 30 at 3:42