Chapter 16 Partial Differentiation

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374 Chapter 16 Partial DifferentiationDEFINITION 16.3.4 Let ∆x = x − x 0 , ∆y = y − y 0 , and ∆z = z − z 0 wherez 0 = f(x 0 ,y 0 ). The function z = f(x,y) is differentiable at (x 0 ,y 0 ) if∆z = f x (x 0 ,y 0 )∆x+f y (x 0 ,y 0 )∆y +ǫ 1 ∆x+ǫ 2 ∆y,and both ǫ 1 and ǫ 2 approach 0 as (x,y) approaches (x 0 ,y 0 ).This definition takes a bit of absorbing. Let’s rewrite the central equation a bit:z = f x (x 0 ,y 0 )(x−x 0 )+f y (x 0 ,y 0 )(y −y 0 )+f(x 0 ,y 0 )+ǫ 1 ∆x+ǫ 2 ∆y. (16.3.1)The first three terms on the right are the equation of the tangent plane, that is,f x (x 0 ,y 0 )(x−x 0 )+f y (x 0 ,y 0 )(y −y 0 )+f(x 0 ,y 0 )is the z-value of the point on the plane above (x,y). Equation 16.3.1 says that the z-valueof a point on the surface is equal to the z-value of a point on the plane plus a “little bit,”namely ǫ 1 ∆x+ǫ 2 ∆y. As (x,y) approaches (x 0 ,y 0 ), both ∆x and ∆y approach 0, so thislittle bit ǫ 1 ∆x + ǫ 2 ∆y also approaches 0, and the z-values on the surface and the planeget close to each other. But that by itself is not very interesting: since the surface andthe plane both contain the point (x 0 ,y 0 ,z 0 ), the z values will approach z 0 and hence getclose to each other whether the tangent plane is “tangent” to the surface or not. The extracondition in the definition says that as (x,y) approaches (x 0 ,y 0 ), the ǫ values approach0—this means that ǫ 1 ∆x+ǫ 2 ∆y approaches 0 much, much faster, because ǫ 1 ∆x is muchsmaller than either ǫ 1 or ∆x. It is this extra condition that makes the plane a tangentplane.We can see that the extra condition on ǫ 1 and ǫ 2 is just what is needed if we look atpartial derivatives. Suppose we temporarily fix y = y 0 , so ∆y = 0. Then the equationfrom the definition becomesor∆z = f x (x 0 ,y 0 )∆x+ǫ 1 ∆x∆z∆x = f x(x 0 ,y 0 )+ǫ 1 .Now taking the limit of the two sides as ∆x approaches 0, the left side turns into thepartial derivative of z with respect to x at (x 0 ,y 0 ), or in other words f x (x 0 ,y 0 ), and theright side does the same, because as (x,y) approaches (x 0 ,y 0 ), ǫ 1 approaches 0. Essentiallythe same calculation works for f y .
16.4 The Chain Rule 375Exercises 16.3.1. Find f x and f y where f(x,y) = cos(x 2 y)+y 3 . ⇒2. Find f x and f y where f(x,y) = xyx 2 +y . ⇒3. Find f x and f y where f(x,y) = e x2 +y 2 . ⇒4. Find f x and f y where f(x,y) = xyln(xy). ⇒5. Find f x and f y where f(x,y) = √ 1−x 2 −y 2 . ⇒6. Find f x and f y where f(x,y) = xtan(y). ⇒7. Find f x and f y where f(x,y) = 1xy . ⇒8. Find an equation for the plane tangent to 2x 2 +3y 2 −z 2 = 4 at (1,1,−1). ⇒9. Find an equation for the plane tangent to f(x,y) = sin(xy) at (π,1/2,1). ⇒10. Find an equation for the plane tangent to f(x,y) = x 2 +y 3 at (3,1,10). ⇒11. Find an equation for the plane tangent to f(x,y) = xln(xy) at (2,1/2,0). ⇒12. Find an equation for the line normal to x 2 +4y 2 = 2z at (2,1,4). ⇒13. Explain in your own words why, when taking a partial derivative of a function of multiplevariables, we can treat the variables not being differentiated as constants.14. Consider a differentiable function, f(x,y). Give physical interpretations of the meanings off x (a,b) and f y (a,b) as they relate to the graph of f.15. In much the same way that we used the tangent line to approximate the value of a functionfrom single variable calculus, we can use the tangent plane to approximate a function frommultivariable calculus. Consider the tangent plane found in Exercise 11. Use this plane toapproximate f(1.98,0.4).16. Suppose that one of your colleagues has calculated the partial derivatives of a given function,and reported to you that f x (x,y) = 2x + 3y and that f y (x,y) = 4x + 6y. Do you believethem? Why or why not? If not, what answer might you have accepted for f y ?17. Suppose f(t) and g(t) are single variable differentiable functions. Find ∂z/∂x and ∂z/∂y foreach of the following two variable functions.a. z = f(x)g(y)b. z = f(xy)c. z = f(x/y)½ºÌ Ò ÊÙÐConsider the surface z = x 2 y +xy 2 , and suppose that x = 2+t 4 and y = 1−t 3 . We canthink of the latter two equations as describing how x and y change relative to, say, time.Thenz = x 2 y +xy 2 = (2+t 4 ) 2 (1−t 3 )+(2+t 4 )(1−t 3 ) 2tells us explicitly how the z coordinate of the corresponding point on the surface dependson t. If we want to know dz/dt we can compute it more or less directly—it’s actually a bit
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Page 19 and 20: 16.5 Directional Derivatives 379So
Page 21 and 22: 16.5 Directional Derivatives 381ori
Page 23 and 24: 16.6 Higher order derivatives 38320
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Chapter 16 Partial Differentiation

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