Lecture Note
University
Massachusetts Institute of TechnologyCourse
Multivariable CalculusPages
2
Academic year
2023
Sporkz
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77
Chain Rule More Variables Let us consider an extension of the chain rule, with an additional variable. π€ = π(π₯, π¦) Let's look at an example in which the quantity w depends on two variables, x and y. xand y depend on a parameter t. However, let us now consider a situation in which xand y themselves depend on two more variables, u and v. π₯ = π₯ π’, π£ ( ) π¦ = π¦(π’, π£) Let us assume that a function is specified in terms of polar coordinates and r andtheta, and then we must switch to the usual coordinates x and y. To express afunction of x and y in terms of the polar coordinates r and theta, first consider theequation for this function. Then we would like to know how the derivatives of afunction with respect to the various sets of variables are related to each other. Oneapproach to this problem is to say that if we substitute the formulas for x and y into f,then w becomes a function of u and v. = π(π₯ π’, π£ ( ), π¦ π’, π£ ( )) If a function can be differentiated, we can find its partial derivatives. If the formulasfor this function are complicated, however, it may not be possible to find their partialderivatives. Normally, if we switch from rectangular to polar coordinates, inversetrigonometric functions may be necessary to express the angle in terms of x and y.And, since we don't want to substitute inverse trigonometric functions everywhere,maybe we'd rather deal with derivatives. How can we do that? Therefore, the question is what are partial w over partial u and partial w over partialof v in terms of. βπ€ βπ’ , βπ€ βπ£ To understand the relationship between w and x, y, you should know how the partialderivatives of w with respect to x and y are related to each other. βπ€ βπ₯ , βπ€ βπ¦ Well, we need to know how x and y depend on u and v. If we do not know that, thenwe do not really know how to do it either. So we also need x sub u, x sub v, y sub u,y sub v. So we have a lot of partials there. Let us start by writing f partial x dx plus f partial y dy. ππ€ = π π₯ ππ₯ + π π¦ ππ¦
Thus far, the differential has been our new friend. Now we'll show how to eliminatedx and dy from an equation in order to express quantities in terms of change in u.How does w change in response to a change in u? First, if we change u a little bit,then x and y will change. How do they change? That is given to us by the differentialdx, which tells us how an increment of u will affect an increment of w. The differentialof a function of two variables is the product of the derivative of each variable withrespect to each other. If that makes sense, then we have the other guy f sub y timeswhat is dy? Well, similarly, dy is y sub u du plus y sub v dv. ππ€ = π π₯ π₯ π’ ππ’ + π₯ π£ ππ£ ( ) + ππ¦(π¦ π’ ππ’ + π¦ π£ ππ£) Now that we have a relation between dw and du and dv, we can express how wreacts to changes in u and v, which was our goal. That is the sum of the product of fsub x times x sub u, plus f sub y times y sub u du, plus f sub x, x sub v, plus f sub y ysub v dv. ππ€ = π π₯ π₯ π’ + π π¦ π¦ π’ ( ) ππ’ + (π π₯ π₯ π£ + π π¦ π¦ π£ )ππ£ Now, dw equals something du plus something dv. The coefficient here must bepartial f partial u. What else could it be? That's the rate of change of w with respectto u if we forget what happens when we change v. That's the definition of a partial. Thus, the total differential accounts for all the partial derivatives that come ascoefficients of individual variables in these expressions.
Chain Rule Expansion: Handling More Variables
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