Midterm SOS 237 package

# Waterloo sos e xam aid math237 e xercise 18 discuss

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Unformatted text preview: t is differentiable at . Thus, gives a good approximation for sufﬁciently close to 17 . WATERLOO SOS E XAM -AID: MATH237 E XERCISE 18. Discuss the validity of the approximation 11 Basic Chain Rule (6.1) and , and and . Let exist, then , and let exists. Moreover, we have the T HEOREM 11.1. Suppose , and . If is differentiable at equation To get an intuition of this theorem, it helps to think of a physical model (a similar example is in your course notes). Consider a metal plate which is heated and cooled non-uniformly. The temperature is a function of and , the position on the plate. The increment form of the linear approximation formula gives the change in corresponding to changes in and : for sufﬁciently small. Suppose we are making measurements at various positions over time; we have , . Dividing by and taking the limit forces the error to disappear and we get the rate of change of the temperature as we measure to be It is important to note that this is not a proof; it is merely a tool for understanding the chain rule. As with other theorems, refer to your course notes for the proof. E XERCISE 19. Show the converse to Theorem 11.1 is not true, i.e. ﬁnd functions such that , and are differentiable at with , and yet is not differentiable at . Hint: Look at a previous exercise. E XAMPLE 21. Suppose S OLUTION. At , we have , We have and . Find at . These are continuous at , so is differentiable at . We also have differentiable at , with 18 WATERLOO SOS E XAM -AID: MATH237 Then by the chain rule, E XERCISE 20. Deﬁne at in two ways. , , . Verify the chain rule theorem by calculating E XERCISE 21. Suppose is differentiable, and deﬁne . Write out the chain rule for . If , calculate . Let . Notation: Recall for a function deﬁned by , and , the chain rule gives In vector notation, this reads or, to be clear about the action of , with deﬁned by . This notation allows us to easily generalize the chain rule to , we have dimensions. For E XERCISE 22. Suppose is differentiable, and deﬁne . Write out the chain rule for . If . Let , calculate . 12 Extensions of the Chain Rule (6.2) E XAMPLE 22. We have shown the chain rule in a very speciﬁc setting, when a function depends on intermediate variables, each of which are dependent on an independent variable . Now suppose there are two independent variables, say where are differentiable. What is the chain rule for ? 19 S OLUTION. Since WATERLOO SOS E XAM -AID: MATH237 is a function of two variables and , we write a chain rule for each of and . It may help to draw a diagram of the chain of dependence. In the previous case of only one independent variable, This diagram represents how is dependent on and , each of which are dependent on the independent variable . Now with two independent variables, we get We can use these dependence diagrams to obtain a chain rule via the following algorithm: (1) Consider all possible paths from the differentiated variable to the differentiating variable. (2) For each link in a given path, differentiate the upper variable with respect to the lower variable. (3) Multiply all derivatives found in step 2. (4) Add all products from step 3, summing over the possible paths. R EMARK 12.1. (a) Remember to hold the correct variables ﬁxed when taking the partial derivatives. For example, means regard as a function of , differentiate with respect to while holding ﬁxed. (b) As before, we need all partial derivatives we take to exist in to guarantee the validity of the chain rule. E XAMPLE 23. Let . be differentiable and let , where and . Compute S OLUTION. The dependence diagram is: 20 WATERLOO SOS E XAM -AID: MATH237 We ca...
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## This note was uploaded on 11/30/2010 for the course MATH 235/237 taught by Professor Wilkie during the Spring '10 term at Waterloo.

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