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3 years ago

Identify the surface with the given vector equation. r(s, t) = s sin 4t, s2, s cos 4t

Mathematics

1 answer:

k0ka [10]3 years ago

4 0

Answer:

Circular paraboloid

Step-by-step explanation:

Given ,

$r(s,t)=ssin4t,s^2,scos4t$

Here, these are the respective $x,y,z$ axes components.

<em>Component along x axis $r_i : ssin4t$ </em>

<em>Component along y axis $r_j : s^2$ </em>

<em>Component along z axis $r_k : scos4t$ </em>

We see that , from the parameterised equation , $r_i^2+r_k^2=s^2sin^24t+s^2cos^24t\\r_i^2+r_k^2=s^2\\r_i^2+r_k^2=r_j$

This can also be written as :

$x^2+z^2=y$

This is similar to an equation of a parabola in 1 Dimension.

By fixing the value of z=0,

<u><em>We get $y=x^2$ which is equation of a parabola curving towards the positive infinity of y-axis and in the x-y plane.</em></u>

By fixing the value of x=0,

<u><em>We get $y=z^2$ which is equation of a parabola curving towards positive infinity of y-axis and in the y-z plane. </em></u>

Thus by fixing the values of x and z alternatively , we get a <u>CIRCULAR PARABOLOID. </u>

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The tread life of tires mounted on light-duty trucks follows the normal probability distribution with a population mean of 60,00

qwelly [4]

Answer:

$S.E=1183.2$

Step-by-step explanation:

From the question we are told that:

Population mean $\mu=60000$

Population standard deviation $\sigma= 14,000 miles.$

Sample size $n=140$

Generally the equation for Standard error is mathematically given by

$S.E=\frac{ \sigma}{ \sqrt n}$

$S.E= \frac{14000}{\sqrt 140}$

$S.E=1183$

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3 years ago

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Answer:

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Step-by-step explanation:

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3 years ago

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Bumek [7]

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3 years ago

18 Which expression shows a "7" as a coefficient? Select ALL that apply.

soldier1979 [14.2K]

Answer:

B,F, C

Step-by-step explanation:

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7 0

3 years ago

find the orthogonal projection of v= [19,12,14,-17] onto the subspace W spanned by [ [ -4,-1,-1,3] ,[ 1,-4,4,3] ] proj w (v) = [

12345 [234]

<h2>Answer:</h2>

Hence, we have:

$proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}]$

<h2>Step-by-step explanation:</h2>

By the orthogonal decomposition theorem we have:

The orthogonal projection of a vector v onto the subspace W=span{w,w'} is given by:

$proj_W(v)=(\dfrac{v\cdot w}{w\cdot w})w+(\dfrac{v\cdot w'}{w'\cdot w'})w'$

Here we have:

$v=[19,12,14,-17]\\\\w=[-4,-1,-1,3]\\\\w'=[1,-4,4,3]$

Now,

$v\cdot w=[19,12,14,-17]\cdot [-4,-1,-1,3]\\\\i.e.\\\\v\cdot w=19\times -4+12\times -1+14\times -1+-17\times 3\\\\i.e.\\\\v\cdot w=-76-12-14-51=-153$

$w\cdot w=[-4,-1,-1,3]\cdot [-4,-1,-1,3]\\\\i.e.\\\\w\cdot w=(-4)^2+(-1)^2+(-1)^2+3^2\\\\i.e.\\\\w\cdot w=16+1+1+9\\\\i.e.\\\\w\cdot w=27$

and

$v\cdot w'=[19,12,14,-17]\cdot [1,-4,4,3]\\\\i.e.\\\\v\cdot w'=19\times 1+12\times (-4)+14\times 4+(-17)\times 3\\\\i.e.\\\\v\cdot w'=19-48+56-51\\\\i.e.\\\\v\cdot w'=-24$

$w'\cdot w'=[1,-4,4,3]\cdot [1,-4,4,3]\\\\i.e.\\\\w'\cdot w'=(1)^2+(-4)^2+(4)^2+(3)^2\\\\i.e.\\\\w'\cdot w'=1+16+16+9\\\\i.e.\\\\w'\cdot w'=42$

Hence, we have:

$proj_W(v)=(\dfrac{-153}{27})[-4,-1,-1,3]+(\dfrac{-24}{42})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=\dfrac{-17}{3}[-4,-1,-1,3]+(\dfrac{-4}{7})[1,-4,4,3]\\\\i.e.\\\\proj_W(v)=[\dfrac{68}{3},\dfrac{17}{3},\dfrac{17}{3},-17]+[\dfrac{-4}{7},\dfrac{16}{7},\dfrac{-16}{7},\dfrac{-12}{7}]\\\\i.e.\\\\proj_W(v)=[\dfrac{464}{21},\dfrac{167}{21},\dfrac{71}{21},\dfrac{-131}{7}]$

6 0

3 years ago