Question: 1 Evaluate the integral [(1 + 3 cost) i + (3 sin t) j+ ( sect) k] at. - 1/4 The result is i +

 1 Evaluate the integral [(1 + 3 cost) i + (3sin t) j+ ( sect) k] at. - 1/4 The result isi + i+ K. (Type exact answers, using x and radicals asneeded.)Solve the initial value problem for r as a vector function of

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t. dr Differential equation: dt = -7ti-t j - 5t k Initialcondition: r(0) = 6i + 8j + 9k r(t) = it kSolvethe initial value problem for r as a vector function of t.dr 9 Differential Equation: E dt 5 (t+1)"i+4elj+ t+ 1 Initial condition:r(0) = k r(t) = + kSolve the initial value problem forr as a vector function of t. 2 dr Differential equation: =

Evaluate the integral [(1 + 3 cost) i + (3 sin t) j+ ( sect) k] at. - 1/4 The result is i + i+ K. (Type exact answers, using x and radicals as needed.)Solve the initial value problem for r as a vector function of t. dr Differential equation: dt = -7ti-t j - 5t k Initial condition: r(0) = 6i + 8j + 9k r(t) = it kSolve the initial value problem for r as a vector function of t. dr 9 Differential Equation: E dt 5 (t+1)"i+4elj+ t+ 1 Initial condition: r(0) = k r(t) = + kSolve the initial value problem for r as a vector function of t. 2 dr Differential equation: = - 4k 2 dt dr Initial conditions: r(0) = 10k and dt = 10i + 10j It= 0 r(t) = + kSolve the following initial value problem for r as a function of t. Differential equation: 2 =2eli-e -tj + 12e 2 k dt Initial conditions: r(0) = 61 + 4j + 6k dr dt = - 2i + 5j It = 0 r(t) = it + kFind the plane determined by the intersecting lines. L1 x= -1+2t y = 2 + 3t z =1-t L2 x=1-4s y = 1+2s z=2-2s III Using a coefficient of - 1 for x, the equation of the plane is (Type an equation.)

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