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borishaifa [10]
2 years ago
8

Find a vector function, r(t), that represents the curve of intersection of the two surfaces. the paraboloid z = 9x2 y2 and the p

arabolic cylinder y = 2x2
Mathematics
1 answer:
dolphi86 [110]2 years ago
7 0

The vector function is, r(t) =  \bold{ < t,2t^2,9t^2+4t^4 > }

Given two surfaces for which the vector function corresponding to the intersection of the two need to be found.

First surface is the paraboloid, z=9x^2+y^2

Second equation is of the parabolic cylinder, y=2x^2

Now to find the intersection of these surfaces, we change these equations into its parametrical representations.

Let x = t

Then, from the equation of parabolic cylinder,  y=2t^2.

Now substituting x and y into the equation of the paraboloid, we get,

z=9t^2+(2t^2)^2 = 9t^2+4t^4

Now the vector function, r(t) = <x, y, z>

So r(t) = \bold{ < t,2t^2,9t^2+4t^4 > }

Learn more about vector functions at brainly.com/question/28479805

#SPJ4

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Determine formula of the nth term 2, 6, 12 20 30,42​
nalin [4]

Check the forward differences of the sequence.

If \{a_n\} = \{2,6,12,20,30,42,\ldots\}, then let \{b_n\} be the sequence of first-order differences of \{a_n\}. That is, for n ≥ 1,

b_n = a_{n+1} - a_n

so that \{b_n\} = \{4, 6, 8, 10, 12, \ldots\}.

Let \{c_n\} be the sequence of differences of \{b_n\},

c_n = b_{n+1} - b_n

and we see that this is a constant sequence, \{c_n\} = \{2, 2, 2, 2, \ldots\}. In other words, \{b_n\} is an arithmetic sequence with common difference between terms of 2. That is,

2 = b_{n+1} - b_n \implies b_{n+1} = b_n + 2

and we can solve for b_n in terms of b_1=4:

b_{n+1} = b_n + 2

b_{n+1} = (b_{n-1}+2) + 2 = b_{n-1} + 2\times2

b_{n+1} = (b_{n-2}+2) + 2\times2 = b_{n-2} + 3\times2

and so on down to

b_{n+1} = b_1 + 2n \implies b_{n+1} = 2n + 4 \implies b_n = 2(n-1)+4 = 2(n + 1)

We solve for a_n in the same way.

2(n+1) = a_{n+1} - a_n \implies a_{n+1} = a_n + 2(n + 1)

Then

a_{n+1} = (a_{n-1} + 2n) + 2(n+1) \\ ~~~~~~~= a_{n-1} + 2 ((n+1) + n)

a_{n+1} = (a_{n-2} + 2(n-1)) + 2((n+1)+n) \\ ~~~~~~~ = a_{n-2} + 2 ((n+1) + n + (n-1))

a_{n+1} = (a_{n-3} + 2(n-2)) + 2((n+1)+n+(n-1)) \\ ~~~~~~~= a_{n-3} + 2 ((n+1) + n + (n-1) + (n-2))

and so on down to

a_{n+1} = a_1 + 2 \displaystyle \sum_{k=2}^{n+1} k = 2 + 2 \times \frac{n(n+3)}2

\implies a_{n+1} = n^2 + 3n + 2 \implies \boxed{a_n = n^2 + n}

6 0
2 years ago
2 Points<br> What is the slope of the line shown below?<br> (2,2) 4,8
Vladimir79 [104]

Answer:

3

Step-by-step explanation:

Slope =y2−y1 divided to x2−x1

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4−2

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= 3

Hope this helped :)

Have a good one

5 0
3 years ago
Read 2 more answers
Will give you brilliantest
Zina [86]
I don’t know but maybe 2
6 0
2 years ago
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