Mixing
Find the equations of the tangent line to the curve
$begingroup$
Given the surface $z = 2x^2 + y^2$:
The plane $y=3$ intersects the surface in a curve. Find the equations of the tangent line to this curve at x = 2.
From my understanding, to get the curve I would plug in $y = 3$, so $z=2x^2 + 9$ is the curve. The tangent line to this curve would be the derivate with respect to $x$, therefor my tangent line is $z = 4x$. And at $x=2$ we simply have $z=8$.
This however is incorrect according to the back of the book as it has the answer as: $y=3$, $z=8x+1$
Could someone please explain to me where my understanding is failing and what would be the correct interpretation?
calculus multivariable-calculus
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
$endgroup$
add a comment |
$begingroup$
Given the surface $z = 2x^2 + y^2$:
The plane $y=3$ intersects the surface in a curve. Find the equations of the tangent line to this curve at x = 2.
From my understanding, to get the curve I would plug in $y = 3$, so $z=2x^2 + 9$ is the curve. The tangent line to this curve would be the derivate with respect to $x$, therefor my tangent line is $z = 4x$. And at $x=2$ we simply have $z=8$.
This however is incorrect according to the back of the book as it has the answer as: $y=3$, $z=8x+1$
Could someone please explain to me where my understanding is failing and what would be the correct interpretation?
calculus multivariable-calculus
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
$endgroup$
$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47
add a comment |
$begingroup$
Given the surface $z = 2x^2 + y^2$:
The plane $y=3$ intersects the surface in a curve. Find the equations of the tangent line to this curve at x = 2.
From my understanding, to get the curve I would plug in $y = 3$, so $z=2x^2 + 9$ is the curve. The tangent line to this curve would be the derivate with respect to $x$, therefor my tangent line is $z = 4x$. And at $x=2$ we simply have $z=8$.
This however is incorrect according to the back of the book as it has the answer as: $y=3$, $z=8x+1$
Could someone please explain to me where my understanding is failing and what would be the correct interpretation?
calculus multivariable-calculus
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
$endgroup$
Given the surface $z = 2x^2 + y^2$:
The plane $y=3$ intersects the surface in a curve. Find the equations of the tangent line to this curve at x = 2.
From my understanding, to get the curve I would plug in $y = 3$, so $z=2x^2 + 9$ is the curve. The tangent line to this curve would be the derivate with respect to $x$, therefor my tangent line is $z = 4x$. And at $x=2$ we simply have $z=8$.
This however is incorrect according to the back of the book as it has the answer as: $y=3$, $z=8x+1$
Could someone please explain to me where my understanding is failing and what would be the correct interpretation?
calculus multivariable-calculus
calculus multivariable-calculus
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
asked Oct 5 '17 at 0:43
FakeBrainFakeBrain
769
769
$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47
add a comment |
$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47
$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47
$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
First off, I am somewhat confused by the statement of the question, so let me try to clarify that, then answer the question that I pose:
The surface in $mathbb{R}^3$ defined by the equation
$$ z = 2x^2 + y^2 $$
intersects the plane $y=3$ in a curve. Find the equations of the line in the plane $y=3$ that is tangent to this curve at $x=2$.
The plane $y=3$ is an isometric copy of the plane ${(x,z):x,zinmathbb{R}}$. Making the obvious identification, this means that we are asked to find a line tangent to the curve
$$ z = 2x^2 + 9 $$
at $x=2$ in the $(x,z)$-plane. The equation of this line is given by
$$ z - z_0 = m(x-x_0), $$
where $(x_0,z_0)$ is a point that the line passes through, and $m$ is the slope of that line.
To get the point, note that we are looking for a line tangent to the curve when $x=2$. When $x=2$, we have $z = 2(2)^2 + 9 = 17$, thus we obtain $(x_0, z_0) = (2,17)$. For the slope, recall that this is given by the derivative. Thus $m$, the slope, is given by
$$ m = left.frac{mathrm{d}z}{mathrm{d}x}right|_{x=2} = left.frac{mathrm{d}}{mathrm{d}x} (2x^2 + 9)right|_{x=2} = left.4xright|_{x=2} = 8. $$
Putting this together, we obtain the equation for the line:
$$ z - 17 = 8(x-2)
implies z = 8x + 1.$$
As pointed out by amd in the comments below, this isn't entirely a complete solution. Up above, we made the identification of the $(x,z)$-plane with the plane $y=3$ in $mathbb{R}^3$. The answer above assumes implicitly that $y=3$. To put the line we found back into $mathbb{R}^3$, we should "undo" this identification. Specifically, the line we are interested in is the set of points
$$ left{ (x,y,z)inmathbb{R}^3 : z=8x+1, y=3 right}.$$
In the context of the original question, I suspect that the equations (note the plural) desired are the two on the right, i.e.
$$ z=8x+1 qquadtext{and}qquad y=3. $$
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
$endgroup$
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
add a comment |
$begingroup$
You are asked to find the tangent to the curve $z = 2x^2+9$ at the point $x = 2$. Note that a tangent line is specified by a slope,and a point on that line. You do not seem to have specified the point, although you have specified the slope.
The slope, indeed, is given by the derivative, which at $x = 2$ is $4x = 8$. That is computed by you.
To find the point, simply substitute $x = 2$ in $z = 2x^2+9$ to get the point $(2,17)$ , which will definitely lie on the tangent line.
Now, the equation of the line in Cartesian coordinates would then be given by $(z - 17) = 8(x - 2)$. This simplifies to $z = 8x+1$, giving you the desired answer.
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
First off, I am somewhat confused by the statement of the question, so let me try to clarify that, then answer the question that I pose:
The surface in $mathbb{R}^3$ defined by the equation
$$ z = 2x^2 + y^2 $$
intersects the plane $y=3$ in a curve. Find the equations of the line in the plane $y=3$ that is tangent to this curve at $x=2$.
The plane $y=3$ is an isometric copy of the plane ${(x,z):x,zinmathbb{R}}$. Making the obvious identification, this means that we are asked to find a line tangent to the curve
$$ z = 2x^2 + 9 $$
at $x=2$ in the $(x,z)$-plane. The equation of this line is given by
$$ z - z_0 = m(x-x_0), $$
where $(x_0,z_0)$ is a point that the line passes through, and $m$ is the slope of that line.
To get the point, note that we are looking for a line tangent to the curve when $x=2$. When $x=2$, we have $z = 2(2)^2 + 9 = 17$, thus we obtain $(x_0, z_0) = (2,17)$. For the slope, recall that this is given by the derivative. Thus $m$, the slope, is given by
$$ m = left.frac{mathrm{d}z}{mathrm{d}x}right|_{x=2} = left.frac{mathrm{d}}{mathrm{d}x} (2x^2 + 9)right|_{x=2} = left.4xright|_{x=2} = 8. $$
Putting this together, we obtain the equation for the line:
$$ z - 17 = 8(x-2)
implies z = 8x + 1.$$
As pointed out by amd in the comments below, this isn't entirely a complete solution. Up above, we made the identification of the $(x,z)$-plane with the plane $y=3$ in $mathbb{R}^3$. The answer above assumes implicitly that $y=3$. To put the line we found back into $mathbb{R}^3$, we should "undo" this identification. Specifically, the line we are interested in is the set of points
$$ left{ (x,y,z)inmathbb{R}^3 : z=8x+1, y=3 right}.$$
In the context of the original question, I suspect that the equations (note the plural) desired are the two on the right, i.e.
$$ z=8x+1 qquadtext{and}qquad y=3. $$
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
$endgroup$
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
add a comment |
$begingroup$
First off, I am somewhat confused by the statement of the question, so let me try to clarify that, then answer the question that I pose:
The surface in $mathbb{R}^3$ defined by the equation
$$ z = 2x^2 + y^2 $$
intersects the plane $y=3$ in a curve. Find the equations of the line in the plane $y=3$ that is tangent to this curve at $x=2$.
The plane $y=3$ is an isometric copy of the plane ${(x,z):x,zinmathbb{R}}$. Making the obvious identification, this means that we are asked to find a line tangent to the curve
$$ z = 2x^2 + 9 $$
at $x=2$ in the $(x,z)$-plane. The equation of this line is given by
$$ z - z_0 = m(x-x_0), $$
where $(x_0,z_0)$ is a point that the line passes through, and $m$ is the slope of that line.
To get the point, note that we are looking for a line tangent to the curve when $x=2$. When $x=2$, we have $z = 2(2)^2 + 9 = 17$, thus we obtain $(x_0, z_0) = (2,17)$. For the slope, recall that this is given by the derivative. Thus $m$, the slope, is given by
$$ m = left.frac{mathrm{d}z}{mathrm{d}x}right|_{x=2} = left.frac{mathrm{d}}{mathrm{d}x} (2x^2 + 9)right|_{x=2} = left.4xright|_{x=2} = 8. $$
Putting this together, we obtain the equation for the line:
$$ z - 17 = 8(x-2)
implies z = 8x + 1.$$
As pointed out by amd in the comments below, this isn't entirely a complete solution. Up above, we made the identification of the $(x,z)$-plane with the plane $y=3$ in $mathbb{R}^3$. The answer above assumes implicitly that $y=3$. To put the line we found back into $mathbb{R}^3$, we should "undo" this identification. Specifically, the line we are interested in is the set of points
$$ left{ (x,y,z)inmathbb{R}^3 : z=8x+1, y=3 right}.$$
In the context of the original question, I suspect that the equations (note the plural) desired are the two on the right, i.e.
$$ z=8x+1 qquadtext{and}qquad y=3. $$
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
$endgroup$
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
add a comment |
$begingroup$
First off, I am somewhat confused by the statement of the question, so let me try to clarify that, then answer the question that I pose:
The surface in $mathbb{R}^3$ defined by the equation
$$ z = 2x^2 + y^2 $$
intersects the plane $y=3$ in a curve. Find the equations of the line in the plane $y=3$ that is tangent to this curve at $x=2$.
The plane $y=3$ is an isometric copy of the plane ${(x,z):x,zinmathbb{R}}$. Making the obvious identification, this means that we are asked to find a line tangent to the curve
$$ z = 2x^2 + 9 $$
at $x=2$ in the $(x,z)$-plane. The equation of this line is given by
$$ z - z_0 = m(x-x_0), $$
where $(x_0,z_0)$ is a point that the line passes through, and $m$ is the slope of that line.
To get the point, note that we are looking for a line tangent to the curve when $x=2$. When $x=2$, we have $z = 2(2)^2 + 9 = 17$, thus we obtain $(x_0, z_0) = (2,17)$. For the slope, recall that this is given by the derivative. Thus $m$, the slope, is given by
$$ m = left.frac{mathrm{d}z}{mathrm{d}x}right|_{x=2} = left.frac{mathrm{d}}{mathrm{d}x} (2x^2 + 9)right|_{x=2} = left.4xright|_{x=2} = 8. $$
Putting this together, we obtain the equation for the line:
$$ z - 17 = 8(x-2)
implies z = 8x + 1.$$
As pointed out by amd in the comments below, this isn't entirely a complete solution. Up above, we made the identification of the $(x,z)$-plane with the plane $y=3$ in $mathbb{R}^3$. The answer above assumes implicitly that $y=3$. To put the line we found back into $mathbb{R}^3$, we should "undo" this identification. Specifically, the line we are interested in is the set of points
$$ left{ (x,y,z)inmathbb{R}^3 : z=8x+1, y=3 right}.$$
In the context of the original question, I suspect that the equations (note the plural) desired are the two on the right, i.e.
$$ z=8x+1 qquadtext{and}qquad y=3. $$
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
$endgroup$
First off, I am somewhat confused by the statement of the question, so let me try to clarify that, then answer the question that I pose:
The surface in $mathbb{R}^3$ defined by the equation
$$ z = 2x^2 + y^2 $$
intersects the plane $y=3$ in a curve. Find the equations of the line in the plane $y=3$ that is tangent to this curve at $x=2$.
The plane $y=3$ is an isometric copy of the plane ${(x,z):x,zinmathbb{R}}$. Making the obvious identification, this means that we are asked to find a line tangent to the curve
$$ z = 2x^2 + 9 $$
at $x=2$ in the $(x,z)$-plane. The equation of this line is given by
$$ z - z_0 = m(x-x_0), $$
where $(x_0,z_0)$ is a point that the line passes through, and $m$ is the slope of that line.
To get the point, note that we are looking for a line tangent to the curve when $x=2$. When $x=2$, we have $z = 2(2)^2 + 9 = 17$, thus we obtain $(x_0, z_0) = (2,17)$. For the slope, recall that this is given by the derivative. Thus $m$, the slope, is given by
$$ m = left.frac{mathrm{d}z}{mathrm{d}x}right|_{x=2} = left.frac{mathrm{d}}{mathrm{d}x} (2x^2 + 9)right|_{x=2} = left.4xright|_{x=2} = 8. $$
Putting this together, we obtain the equation for the line:
$$ z - 17 = 8(x-2)
implies z = 8x + 1.$$
As pointed out by amd in the comments below, this isn't entirely a complete solution. Up above, we made the identification of the $(x,z)$-plane with the plane $y=3$ in $mathbb{R}^3$. The answer above assumes implicitly that $y=3$. To put the line we found back into $mathbb{R}^3$, we should "undo" this identification. Specifically, the line we are interested in is the set of points
$$ left{ (x,y,z)inmathbb{R}^3 : z=8x+1, y=3 right}.$$
In the context of the original question, I suspect that the equations (note the plural) desired are the two on the right, i.e.
$$ z=8x+1 qquadtext{and}qquad y=3. $$
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
edited Oct 5 '17 at 2:05
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
answered Oct 5 '17 at 0:58
Xander HendersonXander Henderson
14.6k103555
14.6k103555
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
add a comment |
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
Your last equation is that of a plane. You need to add the equation $y=3$ to get the line.
$endgroup$
– amd
Oct 5 '17 at 1:57
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
$begingroup$
@amd I had already made the simplification that the only plane that I was working in was the $y=3$ plane (with the obvious identification). However, it seems that I was not clear enough about that, so I'll try to clarify.
$endgroup$
– Xander Henderson
Oct 5 '17 at 1:59
add a comment |
$begingroup$
You are asked to find the tangent to the curve $z = 2x^2+9$ at the point $x = 2$. Note that a tangent line is specified by a slope,and a point on that line. You do not seem to have specified the point, although you have specified the slope.
The slope, indeed, is given by the derivative, which at $x = 2$ is $4x = 8$. That is computed by you.
To find the point, simply substitute $x = 2$ in $z = 2x^2+9$ to get the point $(2,17)$ , which will definitely lie on the tangent line.
Now, the equation of the line in Cartesian coordinates would then be given by $(z - 17) = 8(x - 2)$. This simplifies to $z = 8x+1$, giving you the desired answer.
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
$endgroup$
add a comment |
$begingroup$
You are asked to find the tangent to the curve $z = 2x^2+9$ at the point $x = 2$. Note that a tangent line is specified by a slope,and a point on that line. You do not seem to have specified the point, although you have specified the slope.
The slope, indeed, is given by the derivative, which at $x = 2$ is $4x = 8$. That is computed by you.
To find the point, simply substitute $x = 2$ in $z = 2x^2+9$ to get the point $(2,17)$ , which will definitely lie on the tangent line.
Now, the equation of the line in Cartesian coordinates would then be given by $(z - 17) = 8(x - 2)$. This simplifies to $z = 8x+1$, giving you the desired answer.
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
$endgroup$
add a comment |
$begingroup$
You are asked to find the tangent to the curve $z = 2x^2+9$ at the point $x = 2$. Note that a tangent line is specified by a slope,and a point on that line. You do not seem to have specified the point, although you have specified the slope.
The slope, indeed, is given by the derivative, which at $x = 2$ is $4x = 8$. That is computed by you.
To find the point, simply substitute $x = 2$ in $z = 2x^2+9$ to get the point $(2,17)$ , which will definitely lie on the tangent line.
Now, the equation of the line in Cartesian coordinates would then be given by $(z - 17) = 8(x - 2)$. This simplifies to $z = 8x+1$, giving you the desired answer.
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
$endgroup$
You are asked to find the tangent to the curve $z = 2x^2+9$ at the point $x = 2$. Note that a tangent line is specified by a slope,and a point on that line. You do not seem to have specified the point, although you have specified the slope.
The slope, indeed, is given by the derivative, which at $x = 2$ is $4x = 8$. That is computed by you.
To find the point, simply substitute $x = 2$ in $z = 2x^2+9$ to get the point $(2,17)$ , which will definitely lie on the tangent line.
Now, the equation of the line in Cartesian coordinates would then be given by $(z - 17) = 8(x - 2)$. This simplifies to $z = 8x+1$, giving you the desired answer.
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
edited Jan 16 at 12:41
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
answered Oct 5 '17 at 0:51
астон вілла олоф мэллбэргастон вілла олоф мэллбэрг
38.7k33476
38.7k33476
add a comment |
add a comment |
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$begingroup$
Even in one dimension, the derivative is not the tangent line. The derivative is the slope of the tangent line. I am also somewhat confused–when you say "the" tangent line, do you mean the line in the plane $y=3$ that is tangent to the curve?
$endgroup$
– Xander Henderson
Oct 5 '17 at 0:47