Vectors Calaulus 1.6

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.   The Derivative of a Vector Valued Function Let r(t) be a vector valued function, then                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 (3t2 i + cos(4t) j + tet k)  =  6t i -4sin(t)j + (et + tet) 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.   Properties of Vector Valued Functions Suppose that v(t) and w(t) are vector valued functions, f(t) is a scalar  function, and c is a real number then 1.  d/dt(v(t) + w(t)) = d/dt(v(t)) + d/dt(w(t)) 2.  d/dt(cv(t)) = c d/dt(v(t)) 3.  d/dt(f(t) v(t)) = f '(t) v(t) + f(t) v'(t) 4.  (v(t) . w(t))' = v'(t) . w(t)+ v(t) . w'(t) 5.  (v(t) x w(t))' = v'(t)  x  w(t)+ v(t) x w'(t) 6.  d/dt(v(f(t))) = v'(f(t)) f '(t) 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,         k2  =  ||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 - 8t3 k dt Solution Just take the integral of each component         ( (sin t)dt i) + ( 2t dt  j) -  ( 8t3 dt k)         =  (-cost + c1)i  + (t2 + c2)j  +  (2t4 + c3)k  Notice that we have introduce three different constants, one for each component.