Question

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

Answer 1

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→