#2 #3 #4 #5 #6 Consider the given vector equation. r(t) = 5eti + 3e-tj, t = 0 Fi

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Consider the given vector equation. r(t) = 5eti + 3e-tj, t = 0 Find r (t). r (t) = Sketch the plane curve together with the position vector r(t) and the tangent vector r (t) for the given value of t. Find the derivative, r (t), of the vector function. r(t) = et3i - j + ln (l + 7t)k r (t) = Find the unit tangent vector T(t) at the point with the given value of the parameter t. r(t) = 4 te-t, arctan t, 8et, t = 0 T(0) = Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = 1 + 6 , y = t3 - t, z = t3 + t; (7, 0, 2) (x(t), y(t), z(t)) = Find parametric equations for the tangent line to the curve with the given parametric equations at the specified point. x = t, y = e-2t, z = 3t - t3; (0, 1, 0) (x(t), y(t), z(t)) = Evaluate the integral. 54(t2i + t + t sin pi tk)dt

Explanation / Answer

1) r'(t)= 5e^t i-3e^-t j; at t=0, r' t =5 i-3 j (2)3t^2*e^(t^3) i +(7/1+7t) k; (3) 4i +8j +8 (4) 3,1,4

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