cal4 - Chapter Four Derivatives 4.1 Derivatives Suppose f...

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4.1 Chapter Four Derivatives 4.1 Derivatives Suppose f is a vector function and t 0 is a point in the interior of the domain of f ( t 0 in the interior of a set S of real numbers means there is an interval centered at t 0 that is a subset of S .). The derivative is defined just as it is for a plain old everyday real valued function, except, of course, the derivative is a vector. Specifically, we say that f is differentiable at t 0 if there is a vector v such that lim [ ( ) ( )] t t h t h t + - = 0 1 0 0 f f v . The vector v is called the derivative of f at t 0 . Now, how would we find such a thing? Suppose f i j k ( ) t a t b t c t = + + . Then 1 0 0 0 0 0 0 0 0 h t h t a t h a t h b t h b t h c t h c t h [ ( ) ( )] ( ) ( ) ( ) ( ) ( ) ( ) f f i j k + - = + - + + - + + - . It should now be clear that the vector function f is differentiable at t 0 if and only if each of the coordinate functions a t b t ( ), ( ), and c t ( ) is. Moreover, the vector derivative v is v i j k = + + '( ) '( ) '( ) . Now we “know” what the derivative of a vector function is, and we know how to compute it, but what is it, really? Let’s see. Let f i j ( ) t t t = + 3 This is, of course, a vector function which describes the graph of the function y x = 3 . Let’s look at the
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4.2 derivative of f at t 0 : v i j = + 3 0 2 t . Convince yourself that the direction of the vector v is the direction tangent to the graph of y x = 3 at the point ( , ) t t 0 0 3 . It is not so clear what we should define to be the tangent to a curve other than a plane curve. Again, vectors come to our rescue. If f is a vector description of a space curve, the direction of the derivative f '( ) t vector is the tangent direction at the point f ( ) t -the derivative f '( ) t is said to be tangent to the curve at f ( ) t . If f ( ) t specifies the position of a particle at time t , then, of course, the derivative is the velocity of the particle, and its length | '( )| f t is the speed Thus the distance the particle travels from time t a = to time t b = is given by the integral of the speed: d t dt a b = | '( )| f . If the particle behaves nicely, this distance is precisely the length of the arc of the curve from f ( ) a to f ( ) b It should be clear what we mean by “behaves nicely”. . For the distance traveled by the particle to be the same as the length of its path, there must be no “backtracking”, or reversing direction. This means we must not allow the velocity to be zero for any t between a and b Example Consider the function r i j ( ) cos sin t t t = + Then the derivative, or velocity, is r i j '( ) sin cos t t t = - + . This vector is indeed tangent to the curve described by r (which we already know to be a circle of radius 1 centered at the origin.) at r ( ) t Note that the scalar product r r ( ) '( ) sin cos sin cos t t t t t t = - + = 0, and so the tangent vector and the vector from the center of the circle to the point on the circle are perpendicular-a well- known fact you learned from Mrs. Turner in 4
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cal4 - Chapter Four Derivatives 4.1 Derivatives Suppose f...

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