А теперь - несколько формул...
Oct. 23rd, 2003 11:23 am![[personal profile]](https://www.dreamwidth.org/img/silk/identity/user.png){kind=small}
yuccaВ продолжение темы об американском образовании - письмо, который разослал один русский профессор из моей альма мачехи своим коллегам. Курс 254 - это Calculus 4 (функции нескольких переменных), его берут в основном студенты технических специальностей.
Dear colleagues,
Could you help me to understand some lines in students' papers, MidTerm 1, Class 254 (the sixth quarter of college mathematics)? They are related to
Problem 4. Show that every line normal to the sphere
x^2 + y^2 + z^2 = 1 passes through the origin.
1. In order for 2*i + 2*j + 2*k = 0 at least one value i, j, k must equal 0 because of this we know that every line normal to the sphere passes through the origin.
2. by the dot product every normal line goes threw [sic -bm] 0, 0, 0 b/c all normal vectors at any sphere must go threw [sic - bm] the origin of that sphere.
3. ...therefore, any line normal to the sphere x^2 + y^2 +z^2 = 1 is also parallel to x^2 + y^2 + z^2. So, all of these normal or parallel lines will pass through the origin.
4. By testing the point (2, 2, 2) it shows that every line normal to the plan[sic - bm] must pass through (0, 0, 0) b/c the line that contains pt. (2, 2, 2) is parallel to the sphere therefore it must also pass through (0, 0, 0).
5. x + y + (2/(2^(1/2))) z must isect [sic - bm] 0, 0, 0 bc its normal to graph at ( .5, .5, 1/(2^(1/2)) )
6. The gradient tells us that on any point on the sphere, it's slope in the x, y, and z directions are the same relative to the origin, therefore it can be inferred that any line normal to the sphere which is centered on (0, 0, 0) would
indeed pass through the origin.
7. tangent plane: 2x*(x - x_o) + 2y*(y - y_o) + 2z*(z - z_o) = 0
Every line on this generic plane regardless of (x_o, y_o, z_o) values has a solution at (0, 0, 0). Therefore they pass through the origin regardless of orientation.
8. since the normal is perpendicular to the surface of a sphere is you subtract the normal from itself you will be let with only unit vector and if there are no points, it must go through the origin N - N = i + j + k
9. all lines start from the origin because they are not otherwise expressed to start at another point
10. Since gradient is (2x, 2y, 2z) any line normal to that would pass through center of sphere any normal line will be a function of x, y, z and therefore all
normal lines will have a (0, 0, 0)
11. the sphere x^2 + y^2 + z^2 = 1 passes through the origin at every
line normal to it
12. grad f = (2x, 2y, 2z); at (0, 0, 0) gradf = (0, 0, 0) - therefore the normal vector passes through the origin, so any normal vector must pass through the origin
13. 2x, 2y, 2z all pass through origin since all of the partial derivatives of the equation are normal to the sphere and pass through origin. thus any other normal lines will contain one of these components forcing it through the origin.
14. since gradient at origin is 0, then all normal lines run the same direction as the unit vectors.
15. For all x = y = z not 0 2x + 2y + 2z = 0 so all normal vectors pass thru origin (0, 0, 0)
16. For every level surface along the (z, y), (x, y), (z, x) plane the line perpendicular to the edge of the sphere increases with f_x, f_y, and f_z which are 2x, 2y, 2z so therefore crosses each level surface at the origin because each derivative in x, y, z direction has the same slope (2) for each direction
17, grad f = 2x*i + 2y*j + 2z*k. To NOT pass through origin would have to be parallel to all these axes
18. 2*i + 2*j + 2*z is on a 2:1 ratio with vector i + j + k. Since this forms a 45^o angle (slope 1/1), and since spere [sic - bm] is centered at origin, the normal vector passes through pt (0, 0, 0)
19. [A parabola is drawn - bm] Using 2D picture and convert to scale in 3D
Your help will be greatly appreciated.
Dear colleagues,
Could you help me to understand some lines in students' papers, MidTerm 1, Class 254 (the sixth quarter of college mathematics)? They are related to
Problem 4. Show that every line normal to the sphere
x^2 + y^2 + z^2 = 1 passes through the origin.
1. In order for 2*i + 2*j + 2*k = 0 at least one value i, j, k must equal 0 because of this we know that every line normal to the sphere passes through the origin.
2. by the dot product every normal line goes threw [sic -bm] 0, 0, 0 b/c all normal vectors at any sphere must go threw [sic - bm] the origin of that sphere.
3. ...therefore, any line normal to the sphere x^2 + y^2 +z^2 = 1 is also parallel to x^2 + y^2 + z^2. So, all of these normal or parallel lines will pass through the origin.
4. By testing the point (2, 2, 2) it shows that every line normal to the plan[sic - bm] must pass through (0, 0, 0) b/c the line that contains pt. (2, 2, 2) is parallel to the sphere therefore it must also pass through (0, 0, 0).
5. x + y + (2/(2^(1/2))) z must isect [sic - bm] 0, 0, 0 bc its normal to graph at ( .5, .5, 1/(2^(1/2)) )
6. The gradient tells us that on any point on the sphere, it's slope in the x, y, and z directions are the same relative to the origin, therefore it can be inferred that any line normal to the sphere which is centered on (0, 0, 0) would
indeed pass through the origin.
7. tangent plane: 2x*(x - x_o) + 2y*(y - y_o) + 2z*(z - z_o) = 0
Every line on this generic plane regardless of (x_o, y_o, z_o) values has a solution at (0, 0, 0). Therefore they pass through the origin regardless of orientation.
8. since the normal is perpendicular to the surface of a sphere is you subtract the normal from itself you will be let with only unit vector and if there are no points, it must go through the origin N - N = i + j + k
9. all lines start from the origin because they are not otherwise expressed to start at another point
10. Since gradient is (2x, 2y, 2z) any line normal to that would pass through center of sphere any normal line will be a function of x, y, z and therefore all
normal lines will have a (0, 0, 0)
11. the sphere x^2 + y^2 + z^2 = 1 passes through the origin at every
line normal to it
12. grad f = (2x, 2y, 2z); at (0, 0, 0) gradf = (0, 0, 0) - therefore the normal vector passes through the origin, so any normal vector must pass through the origin
13. 2x, 2y, 2z all pass through origin since all of the partial derivatives of the equation are normal to the sphere and pass through origin. thus any other normal lines will contain one of these components forcing it through the origin.
14. since gradient at origin is 0, then all normal lines run the same direction as the unit vectors.
15. For all x = y = z not 0 2x + 2y + 2z = 0 so all normal vectors pass thru origin (0, 0, 0)
16. For every level surface along the (z, y), (x, y), (z, x) plane the line perpendicular to the edge of the sphere increases with f_x, f_y, and f_z which are 2x, 2y, 2z so therefore crosses each level surface at the origin because each derivative in x, y, z direction has the same slope (2) for each direction
17, grad f = 2x*i + 2y*j + 2z*k. To NOT pass through origin would have to be parallel to all these axes
18. 2*i + 2*j + 2*z is on a 2:1 ratio with vector i + j + k. Since this forms a 45^o angle (slope 1/1), and since spere [sic - bm] is centered at origin, the normal vector passes through pt (0, 0, 0)
19. [A parabola is drawn - bm] Using 2D picture and convert to scale in 3D
Your help will be greatly appreciated.
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Еше одна формула
Date: 2003-10-24 12:31 am (UTC)"When I was calculus czar, I wanted to start to get some idea of what our students were learning from us in calculus."
Так что он спросил студентов, которые уже "взяли" весь calculus и "берут" следуюший курс (инженера будущие):
"What kind of series is 3 + 3r + 3r^2 + 3r^3 + ... & and where, if anyplace, does this converge? If it converges, what is its sum?
A common response was: It is an infinite series and it doesn't converge anyplace."