Midterm SOS 237 package

# To nd there are two paths along the path and along

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Unformatted text preview: n then use our algorithm to ﬁnd the partial derivatives. To ﬁnd , there are two paths: along the path and along the path. We can make a similar argument for the partial derivatives with respect to and , and we get: Then E XAMPLE 24. We can also have a function that depends directly on the independent variable, say where for ? are differentiable, and the partial derivative of with respect to exists. What is the chain rule S OLUTION. The dependence diagram is Then by our algorithm, Notice is the ordinary derivative of as a composite function of , while as a function of , with ﬁxed. To avoid confusion, we recall is the partial derivative of and we might write 21 WATERLOO SOS E XAM -AID: MATH237 E XAMPLE 25. What happens if we have a function that depends directly on two independent variables? For example, suppose where all functions are differentiable. What is the chain rule for ? S OLUTION. The dependence diagram is Then by our algorithm, In paticular, notice we have the symbol appearing on both sides of the equation - do not get confused by this! On the left hand side, we are considering as the composite function of , and taking the partial derivative with respect to . On the right hand side, we are considering as a function of , and holding ﬁxed. To avoid confusion, we recall and we might write E XERCISE 23. Suppose on must be made? , with. Let . Calculate . What assumptions E XAMPLE 26. Thus far we have only dealt with chain rule for the ﬁrst (partial) derivative. Suppose , with , , and is twice differentiable. Find the chain rule for . S OLUTION. To ﬁnd the chain rule for the second derivative, we follow the same type of algorithm. To begin, we must calculate the ﬁrst (partial) derivative using the chain rule. The dependence diagram is Then by our algorithm, Then is dependent on , where are each dependent on 22 . We get the dependence diagram WATERLOO SOS E XAM -AID: MATH237 Then by our algorithm, We then calculate in the usual way. Since is ﬁxed, Similarly, Finally, holding ﬁxed gives Then E XERCISE 24. Suppose . Find is twice differentiable. Deﬁne . for constants 13 Directional Derivatives (7.1) In this section, we generalize our notion of partial derivative. Recall was the rate of change in along the -direction, while was the rate of change in along the -direction. To get more information about the behaviour of , we would like to consider how it changes along arbitrary lines. D EFINITION 13.1. The directional derivative of deﬁned by at a point in the direction of a unit vector is provided this limit exists. Notice if is differentiable at , this becomes 23 WATERLOO SOS E XAM -AID: MATH237 For 2 dimensions consider the intersection of the vertical plane parallel to the vector at , with the surface . Geometrically, the directional derivative is the slope of the tangent of this intersection at . the point R EMARK 13.2. (a) If we choose or partial derivatives and respectively. (b) Remember that to apply the deﬁnition, we require , then the directional derivative is the to be a unit vector. E XAMPLE 27. Deﬁne in the direction S OLUTION. Since by . . Find the directional derivative of at the point is not a unit vector, we must ﬁrst normalize: Then since is differentiable, There is an easier way to calculate directional derivatives, if is differentiable. If it is not, you must go back to the limit deﬁnition, be careful of this (Recall the partial derivatives can exist and the function not be differentiable). T HEOREM 13.3. If is differentiable at , then E XAMPLE 28. Deﬁne in the direction S OLUTION. Since by using Theorem 13.3. . Find the directional derivative of a...
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