3 the sides of a triangle vary in such a way that the

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3. The sides of a triangle vary in such a way that the area remains constant, so that a may be regarded as a function of b and c . Prove that ∂a ∂b = - cos B cos A , ∂a ∂c = - cos C cos A . [This follows from the equations da = ∂a ∂b db + ∂a ∂c dc, cos A da + cos B db + cos C dc = 0 . ] 4. If a , b , c vary so that R remains constant, then da cos A + db cos B + dc cos C = 0 , and so ∂a ∂b = - cos A cos B , ∂a ∂c = - cos A cos C . [Use the formulae a = 2 R sin A , . . . , and the facts that R and A + B + C are constant.] 5. If z is a function of u and v , which are functions of x and y , then ∂z ∂x = ∂z ∂u ∂u ∂x + ∂z ∂v ∂v ∂x , ∂z ∂y = ∂z ∂u ∂u ∂y + ∂z ∂v ∂v ∂y .
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[VII : 155] ADDITIONAL THEOREMS IN THE CALCULUS 346 [We have dz = ∂z ∂u du + ∂z ∂v dv, du = ∂u ∂x dx + ∂u ∂y dy, dv = ∂v ∂x dx + ∂v ∂y dy. Substitute for du and dv in the first equation and compare the result with the equation dz = ∂z ∂x dx + ∂z ∂y dy. ] 6. Let z be a function of x and y , and let X , Y , Z be defined by the equations x = a 1 X + b 1 Y + c 1 Z, y = a 2 X + b 2 Y + c 2 Z, z = a 3 X + b 3 Y + c 3 Z. Then Z may be expressed as a function of X and Y . Express ∂Z/∂X , ∂Z/∂Y in terms of ∂z/∂x , ∂z/∂y . [Let these differential coefficients be denoted by P , Q and p , q . Then dz - p dx - q dy = 0, or ( c 1 p + c 2 q - c 3 ) dZ + ( a 1 p + a 2 q - a 3 ) dX + ( b 1 p + b 2 q - b 3 ) dY = 0 . Comparing this equation with dZ - P dX - Q dY = 0 we see that P = - a 1 p + a 2 q - a 3 c 1 p + c 2 q - c 3 , Q = - b 1 p + b 2 q - b 3 c 1 p + c 2 q - c 3 . ] 7. If ( a 1 x + b 1 y + c 1 z ) p + ( a 2 x + b 2 y + c 2 z ) q = a 3 x + b 3 y + c 3 z, then ( a 1 X + b 1 Y + c 1 Z ) P + ( a 2 X + b 2 Y + c 2 Z ) Q = a 3 X + b 3 Y + c 3 Z. ( Math. Trip. 1899.) 8. Differentiation of implicit functions. Suppose that f ( x, y ) and its derivative f 0 y ( x, y ) are continuous in the neighbourhood of the point ( a, b ), and that f ( a, b ) = 0 , f 0 b ( a, b ) 6 = 0 .
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[VII : 156] ADDITIONAL THEOREMS IN THE CALCULUS 347 Then we can find a neighbourhood of ( a, b ) throughout which f 0 y ( x, y ) has always the same sign. Let us suppose, for example, that f 0 y ( x, y ) is positive near ( a, b ). Then f ( x, y ) is, for any value of x sufficiently near to a , and for values of y sufficiently near to b , an increasing function of y in the stricter sense of § 95 . It follows, by the theorem of § 108 , that there is a unique continuous function y which is equal to b when x = a and which satisfies the equation f ( x, y ) = 0 for all values of x sufficiently near to a . Let us now suppose that f ( x, y ) possesses a derivative f 0 x ( x, y ) which is also continuous near ( a, b ). If f ( x, y ) = 0, x = a + h , y = b + k , we have 0 = f ( x, y ) - f ( a, b ) = ( f 0 a + ) h + ( f 0 b + η ) k, where and η tend to zero with h and k . Thus k h = - f 0 a + f 0 b + η → - f 0 a f 0 b , or dy dx = - f 0 a f 0 b . 9. The equation of the tangent to the curve f ( x, y ) = 0, at the point x 0 , y 0 , is ( x - x 0 ) f 0 x 0 ( x 0 , y 0 ) + ( y - y 0 ) f 0 y 0 ( x 0 , y 0 ) = 0 . 156. Definite Integrals and Areas. It will be remembered that, in Ch. VI , § 145 , we assumed that, if f ( x ) is a continuous function of x , and PQ is the graph of y = f ( x ), then the region PpqQ shown in Fig. 47 has associated with it a definite number which we call its area . It is clear that, if we denote Op and Oq by a and x , and allow x to vary, this area is a function of x , which we denote by F ( x ).
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