Differentiation and Integration of Vector Valued Functions Calculus of Vector Valued Functions The formal definition of the derivative of a vector valued function is very similar to the definition of the derivative of a real valued function.
Because the derivative of a sum is the sum of the derivative, we can find the derivative of each of the components of the vector valued function to find its derivative.
Examples d/dt (3i + sintj) = costj d/dt (3t^{2} i + cos(4t) j + te^{t} k) = 6t i 4sin(t)j + (e^{t} + te^{t}) k
Properties of Vector Valued Functions All of the properties of differentiation still hold for vector values functions. Moreover because there are a variety of ways of defining multiplication, there is an abundance of product rules.
Example Show that if r is a differentiable vector valued function with constant magnitude, then r ^{.} r' = 0
Solution Since r has constant magnitude, call it k, k^{2} = r^{2} = r ^{.} r Taking derivatives of the left and right sides gives 0 = (r ^{.} r)' = r' ^{.} r + r ^{.} r' = r ^{.} r' + r ^{.} r' = 2r ^{.} r' Divide by two and the result follows Integration of vector valued functions We define the integral of a vector valued function as the integral of each component. This definition holds for both definite and indefinite integrals.
Example Evaluate (sin t)i + 2t j  8t^{3 }k dt
Solution Just take the integral of each component ( (sin t)dt i) + ( 2t dt j)  ( 8t^{3} dt k) = (cost + c_{1})i + (t^{2} + c_{2})j + (2t^{4} + c_{3})k
Notice that we have introduce three different constants, one for each component.
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