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Bumek [7]
3 years ago
13

Please help. I need an answer ASAP. Calculus question.

Mathematics
1 answer:
Nataly [62]3 years ago
6 0

Answer:

r(t) =2/3t^3 +3t +1 i→  - 1/20 t^5  +t +1 j→  + e^-t -t -2 k→

Step-by-step explanation:

Acceleration is the derivative of velocity,  We will need to integrate acceleration to get the velocity.  Remembering to keep them as vectors

∫a dt  =∫ 4x   i→ - x^3   j→ +e ^-x  k→) dt  from 0 to t

        v(x) = 4x^2  /2  i→ -x^4  /4  j→  - e^-x k→  +c  from 0 to t  

      v(t)    =4t^2  /2  i→ -t^4  /4  j→  - e^-t k→  +c    

      v(t)    =2t^2    i→ - 1/4t^4   j→  - e^-t k→  +c  

           where c is the constant of integration

We can determine c using v(0)

      v(t)    =2t^2  i→   -1/4t^4    j→  - e^-t k→  +c  

Substituting t= 0

v(0) =   2(0) i→   - 1/4 *0   j→  - e^-0 k→  +c  

We know v(0)= 3 i→   + 1   j→  - 2 k→

Setting them equal

3 i→   + 1   j→  - 2 k→ =   2(0) i→   - 1/4 *0   j→  - e^-0 k→  +c  

3 i→   + 1   j→  - 2 k→ =     -1 k→  +c  

Add 1k to each side

3 i→   + 1   j→ 1k - 2 k→ =    1k+ -1 k→  +c  

3 i→   +1   j→ -1 k→ =   c  

 v(t)    =2t^2    i→ - 1/4t^4   j→  - e^-t k→ +   3 i→   + 1   j→ -1 k→

Combining

v(t) = 2t^2 +3   i→ - 1/4t^4+1   j→  - e^-t  -1k→  

Now we need  to integrate again to get d(t).  Velocity is the derivative of position  We will need to integrate acceleration to get the position.  Remembering to keep them as vectors

∫v dt  =∫( 2x^2 +3   i→ - 1/4x^4+1   j→  - e^-x  -1k→) dt  from 0 to t

r(t)   = 2 x^3/ 3 +3x  i→  - 1/4 x^5 /5 +x  j→  + e^-x -x k→ +c2  where c2 is the constant of integration

r(t)   =  2/3t^3 +3t  i→  - 1/20 t^5  +t  j→  + e^-t -tk→  +c2

We can determine c2 using r(0)

      r(0)    = 2/3t^3 +3t  i→  - 1/20 t^5  +t  j→  + e^-t -t k→ +c2

Substituting t=0

  r(0)    = 2/3 *0^3 +3*0  i→  - 1/20 *0^5  +0  j→  + e^-0 -0 k→ +c2

We know r(0)= 1 i→   + 1   j→  - 1 k→

Setting them equal

1 i→   + 1   j→  - 1 k→  = 2/3 *0^3 +3*0  i→  - 1/20 *0^5  +0  j→  + e^-0 -0  k→+c2

1 i→   + 1   j→  - 1 k→ =  0  i→  - 0  j→  + 1 k→ +c2

1 i→   + 1   j→  - 1 k→  = 1 k→ +c2

Subtract  1 k→ from each side

1 i→   + 1   j→  - 1 k -1 k→=   1 k→  1 k→ +c2

1 i→   + 1   j→  - 1 k -1 k→ =  1 k→  1 k→ +c2

1 i→   + 1   j→  - 2 k→ =   c2

r(t) = 2/3t^3 +3t  i→  - 1/20 t^5  +t  j→  + e^-t -tk→  +c2

r(t) =2/3t^3 +3t  i→  - 1/20 t^5  +t  j→  + e^-t -tk→   1 i→   + 1   j→  - 2 k→

Combining

r(t) =2/3t^3 +3t +1 i→  - 1/20 t^5  +t +1 j→  + e^-t -t -2 k→

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Answer:

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Step-by-step explanation:

Order the integers from least to greatest which means that the smallest number to the biggest number. Since the biggest number is 100 it goes at the end and 0 is the smallest number so it goes at the end. After that, all you have to do is find which numbers are bigger and smaller. So, the numbers should go in this order 0, 5, 18, 22, 100.

Hope This Helps :)

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