query_10_2

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course Mth 277

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Question: Find both F' and F'' for F(t) = (4sin^2 t)i + (9cos^2 t)j + tk

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Your solution:

F'=8sin(t)cos(t)-18cos(t)sin(t)

F'=4sin(2t)i-9sin(2t)j+k

F""=8cos(2t)i-18cos(2t)j

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Given the position vector of a particle R(t) = (cos t)i + tj + (4 sin t)k, find the particle's velocity and acceleration vectors and then find the speed and direction of the particle at t = pi/2.

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Your solution:

R(t)=(cost)i+tj+(4sint)k

V(t)=-sin(t)i+j+4cos(t)k

a(t)=-cos(t)i-4sin(t)k

speed is the magn of the vel

magn V(pi/2)=sqrt(-1^2+1^2+0^2)=sqrt(2)rad/s

direction=-i/sqrt(2)+j/sqrt(2)

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find Int( dt) (Where Int( f(t) dt) is the integral of f with respect to t)

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Your solution:

int( dt)=<-cos(t),sin(t),t^3/3>

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: Find Integral((e^t)* dt)

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Your solution:

Integral((e^t)* dt)

Integral(( dt)

confidence rating #$&*:

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Given Solution:OK

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Self-critique (if necessary):OK

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Self-critique rating:

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Question: Find the velocity and position vectors given the acceleration vector `A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k, initial position R(0) = 2i + j -3k and initial velocity v(0) = 4i + j + 2k.

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Your solution:

A(t) = 4(t^2)i - 2 sqrt(t) j + 5(e^3t)k

Int(A(t))=V(t)

V(t)=((4/3)(t^3)+c)i-((4/3)t^(3/2)+c)j+(5/3(e^3t)+c)k

V(0)=c1i-c2j+(5/3+c3)k

c1=4

c2=-1

c3=1/3

V(t)=V(t)=((4/3)(t^3)+4)i-((4/3)t^(3/2)-1)j+(5/3(e^3t)+1/3)k

R(t)=Int(V(t))

R(t)=((1/3)t^4+4t+c)i-((8/15)t^(5/2)-t+c)j+(5/9(e^3t)+(1/3)t+c)k

R(0)=c1i-c2j+(5/9+c3)k

c1=2

c2=-1

c3=-32/9

R(t)=((1/3)t^4+4t+2)i-((8/15)t^(5/2)-t-1)j+(5/9(e^3t)+(1/3)t-32/9)k

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel.

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Your solution:

F(t)=e^(-kt)i+e^(kt)k

F'(t)=-ke^(-kt)i+ke^(kt)k

F""(t)=(k^2)e^(-kt)i+(k^2)e^(kt)k

Yes they are parallel because they are multiples of eachother. Even though we do not know the value of the constant k it,k^2, is multiplied through both components and the rest of the equation is the same so they are parallel.

confidence rating #$&*:

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Given Solution:

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Self-critique (if necessary):OK

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Self-critique rating:OK

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Question: "

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Question: "

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#*&!

query_10_2

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