PureMath.pdf

# Say by an increment mm 1 δx while keeping y constant

This preview shows pages 352–356. Sign up to view the full content.

, say by an increment MM 1 = δx , while keeping y constant. This brings P to P 1 . If along OP 1 we take OP 0 = OP , the increment of r is P 0 P 1 = δr , say; and ∂r/∂x = lim( δr/δx ). If on the other hand we want to calculate ∂x/∂r , x and y being now regarded as functions of r and θ , we must increase r by Δ r , say, keeping θ constant. This brings P to P 2 , where PP 2 = Δ r : the corresponding increment of x is MM 1 = Δ x , say; and ∂x/∂r = lim(Δ x/ Δ r ) .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
[VII : 152] ADDITIONAL THEOREMS IN THE CALCULUS 337 O M M M 1 P P P 1 P 2 Fig. 46. Now Δ x = δx : * but Δ r 6 = δr . Indeed it is easy to see from the figure that lim( δr/δx ) = lim( P 0 P 1 /PP 1 ) = cos θ, but lim(Δ r/ Δ x ) = lim( PP 2 /PP 1 ) = sec θ, so that lim( δr/ Δ r ) = cos 2 θ. The fact is of course that ∂x/∂r and ∂r/∂x are not formed upon the same hypothesis as to the variation of P . ] 3. Prove that if z = f ( ax + by ) then b ( ∂z/∂x ) = a ( ∂z/∂y ). 4. Find ∂X/∂x , ∂X/∂y , . . . when X + Y = x , Y = xy . Express x , y as functions of X , Y and find ∂x/∂X , ∂x/∂Y , . . . . 5. Find ∂X/∂x , . . . when X + Y + Z = x , Y + Z = xy , Z = xyz ; express x , y , z in terms of X , Y , Z and find ∂x/∂X , . . . . [There is of course no difficulty in extending the ideas of the last section to functions of any number of variables. But the reader must be careful to * Of course the fact that Δ x = δx is due merely to the particular value of Δ r that we have chosen (viz. PP 2 ). Any other choice would give us values of Δ x , Δ r proportional to those used here.
[VII : 153] ADDITIONAL THEOREMS IN THE CALCULUS 338 impress on his mind that the notion of the partial derivative of a function of several variables is only determinate when all the independent variables are specified. Thus if u = x + y + z , x , y , and z being the independent variables, then ∂u/∂x = 1. But if we regard u as a function of the variables x , x + y = η , and x + y + z = ζ , so that u = ζ , then ∂u/∂x = 0.] 153. Differentiation of a function of two functions. There is a theorem concerning the differentiation of a function of one variable, known generally as the Theorem of the Total Differential Coefficient , which is of very great importance and depends on the notions explained in the preceding section regarding functions of two variables. This theorem gives us a rule for differentiating f { φ ( t ) , ψ ( t ) } , with respect to t . Let us suppose, in the first instance, that f ( x, y ) is a function of the two variables x and y , and that f 0 x , f 0 y are continuous functions of both variables ( § 107 ) for all of their values which come in question. And now let us suppose that the variation of x and y is restricted in that ( x, y ) lies on a curve x = φ ( t ) , y = ψ ( t ) , where φ and ψ are functions of t with continuous differential coefficients φ 0 ( t ), ψ 0 ( t ). Then f ( x, y ) reduces to a function of the single variable t , say F ( t ). The problem is to determine F 0 ( t ). Suppose that, when t changes to t + τ , x and y change to x + ξ and y + η . Then by definition dF ( t ) dt = lim τ 0 1 τ [ f { φ ( t + τ ) , ψ ( t + τ ) } - f { φ ( t ) , ψ ( t ) } ] = lim 1 τ { f ( x + ξ, y + η ) - f ( x, y ) } = lim f ( x + ξ, y + η ) - f ( x, y + η ) ξ ξ τ + f ( x, y + η ) - f ( x, y ) η η τ .

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
[VII : 153] ADDITIONAL THEOREMS IN THE CALCULUS 339 But, by the Mean Value Theorem, { f ( x + ξ, y
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern