#$&* course Mth 277 query_10_2*********************************************
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Given the position vector of a particle R(t) = (cos t)i + tj + (4 sin t)k, find the particle's velocity and acceleration vectors and then find the speed and direction of the particle at t = pi/2. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: R(t)=(cost)i+tj+(4sint)k V(t)=-sin(t)i+j+4cos(t)k a(t)=-cos(t)i-4sin(t)k speed is the magn of the vel magn V(pi/2)=sqrt(-1^2+1^2+0^2)=sqrt(2)rad/s direction=-i/sqrt(2)+j/sqrt(2) confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find Int(
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: Find Integral((e^t)*
.............................................
Given Solution:OK &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating: ********************************************* 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. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY 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 #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: F(t) = e^(-kt)i + e^(kt)k. Show that F and F'' are parallel. YYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYYY Your solution: F(t)=e^(-kt)i+e^(kt)k F'(t)=-ke^(-kt)i+ke^(kt)k F""(t)=(k^2)e^(-kt)i+(k^2)e^(kt)k Yes they are parallel because they are multiples of eachother. Even though we do not know the value of the constant k it,k^2, is multiplied through both components and the rest of the equation is the same so they are parallel. confidence rating #$&*: ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
.............................................
Given Solution: &&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&&& Self-critique (if necessary):OK ------------------------------------------------ Self-critique rating:OK ********************************************* Question: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: ********************************************* Question: " Self-critique (if necessary): ------------------------------------------------ Self-critique rating: #*&!