Find the sum of the given vectors and illustrate geometrically. [0,1,2],[0,0,-3]

Mathematics

1 answer:

Nitella [24]3 years ago

5 0

Answer:

The sum of the given vectors is $[0,1,2]+[0,0,-3]=[0,1,-1]$

Step-by-step explanation:

For the given vectors (which are R³ vectors), the sum is simply the sum of each coordinate, if a general vector is written as

$[x,y,z]$

the sum of two vector will be in each coordinate at a time.

To illustrate geometrically the resulting vector in the space xyz

$[0,1,-1]$

we can say that the first coordinate is on the x-axis, the sencond on the y-axis, and the third one on the z-axis, so the illustration will be a vector starting from the center of coordinates, and ending in the coordinates 0 of the x-axis, 1 of the y-axis, and -1 of the z-axis. Or, in a plane yz (where x=0), a vector from the origin to the point 1 in y-axis, and -1 in z-axis.

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How do you find the volume of the solid generated by revolving the region bounded by the graphs

d1i1m1o1n [39]

Answer:

About the x axis

$V = 4\pi[ \frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

$V = \pi [4y -y^2 +\frac{y^3}{12}] \Big|_0^8 =\pi *\frac{32}{3}= \frac{32 \pi}{3}$

About the line y=8

$V = \pi [64x -\frac{32}{3}x^3 +\frac{4}{5}x^5] \Big|_0^2 =\pi *(128-\frac{256}{3} +\frac{128}{5})= \frac{1024 \pi}{5}$

About the line x=2

$V = \frac{\pi}{2} [\frac{y^2}{2}] \Big|_0^8 =\frac{\pi}{4} *(64)= 16\pi$

Step-by-step explanation:

For this case we have the following functions:

$y = 2x^2 , y=0, X=2$

About the x axis

Our zone of interest is on the figure attached, we see that the limit son x are from 0 to 2 and on y from 0 to 8.

We can find the area like this:

$A = \pi r^2 = \pi (2x^2)^2 = 4 \pi x^4$

And we can find the volume with this formula:

$V = \int_{a}^b A(x) dx$

$V= 4\pi \int_{0}^2 x^4 dx$

$V = 4\pi [\frac{x^5}{5}] \Big|_0^2 =4\pi *\frac{32}{5}= \frac{128 \pi}{5}$

About the y axis

For this case we need to find the function in terms of x like this:

$x^2 = \frac{y}{2}$

$x = \pm \sqrt{\frac{y}{2}}$ but on this case we are just interested on the + part $x=\sqrt{\frac{y}{2}}$ as we can see on the second figure attached.