UNIVERSIDAD CARLOS III DE MADRID MATHEMATICS II EXERCISES (SOLUTIONS ) CHAPTER 3: Partial derivatives and differentiation

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1 UNIVERSIDAD CARLOS III DE MADRID MATHEMATICS II EXERCISES SOLUTIONS ) 3-1. Find, for the following functions: a) fx, y) x cos x sin y. b) fx, y) e xy. c) fx, y) x + y ) lnx + y ). CHAPTER 3: Partial derivatives and differentiation a) The partial derivatives of the function fx, y) xcos x)sin y) are x, y) cos x sin y x sin x sin y, b) The partial derivatives of the function fx, y) e xy are x, y) y e xy, x, y) x, y) xye xy x cos x cos y c) The partial derivatives of the function fx, y) x + y ) lnx + y ) are x, y) x ln x + y ) + x, x, y) y ln x + y ) + y 3-. Determine the marginal-products for the following Cobb-Douglas production function. F x, y, z) 1x 1/ y 1/3 z 1/4 We compute the partial derivatives with respect to each factor, F F F 6x 1/ y 1/3 z 1/4 4x 1/ y /3 z 1/4 3x 1/ y 1/3 z 3/ Find the gradient of the following functions at the given point p a) fx, y) a x y ) 1/ at p a/, a/). b) gx, y) ln1 + xy) 1/ at p 1, 1). c) hx, y) e y cos3x + y) at p π/3, 0). ) a) a x y ) 1/ x, y so the gradient is a x y ) a x y ) b) ln1 + xy) 1/ ) 1 c) x a x y ), y 1, a x y )) 1 ) xya/ ) y 1+xy, x 1+xy so the gradient is 1 4, 4) 1. e y cos3x + y)) xπ/3,y0 3e y sin 3x + y), e y cos 3x + y) e y sin 3x + y)) xπ/3,y0 0, 1) 1

2 3-4. Consider the function fx, y) { x +y x + y sinxy) if x, y) 0, 0) 0 if x, y) 0, 0) a) Find the partial derivatives of f at the point 0, 0). b) Prove that f is continuous on all of R. Hint: Use proving it) that for x, y) 0, 0) we have that x + y 0 1 x + y c) Is f differentiable at 0, 0)? a) The partial derivative with respect to x is since sin0) 0. Likewise, so, ft, 0) f0, 0) 0, 0) lim 0 t 0 t f0, t) f0, 0) 0, 0) lim 0 t 0 t f0, 0) 0, 0) b) Note that for x, y) 0, 0), the function f is a quotient of continuous functions and, hence is continuous for x, y) 0, 0). Recall that the function f is continuous at the point 0, 0) if lim fx, y) f0, 0) x,y) 0,0) We prove this next. First, since x, y 0 then Hence, x + y x + y + x y x + y ) x + y 0 1) x + y Given ε > 0, we take δ ε/. If 0 < x + y < δ then fx, y) x + y sinxy) since sinxy) 1) x + y x + y x + y x + y x + y x + y x + y < δ ε by the above observation) c) First, we note that f is differentiable in R \ {0, 0)}, because the partial derivatives exist and are continuous at every point of R \ {0, 0)}. The function is differentiable at 0, 0) if fv 1, v ) f0, 0) f0, 0) v 1, v ) lim 0 v 1,v ) 0,0) v 1 + v Note that f0, 0) 0, f0, 0) v 1, v ) 0. So, let us consider the quotient fv 1, v ) v1 + v v v 1 + v v 1 + v ) v1 + sinv 1 v ) 1 + v v v 1 + v ) sinv 1v ) con v 1, v ) 0, 0). The function f is differentiable at 0, 0) if v lim 1 + v v 1,v ) 0,0) v 1 + v ) sinv 1v ) 0

3 3 By the observation made in the previous part, we have that v v v 1 + v ) sinv 1v ) sinv 1 v ) and since sinv 1 v ) is continuous, Hence, and f is differentiable at 0, 0). 3-. Consider the function lim v 1,v ) 0,0) fx, y) lim sinv 1v ) 0 v 1,v ) 0,0) v 1 + v v 1 + v ) sinv 1v ) 0 { x sin y x +y ifx, y) 0, 0) 0 ifx, y) 0, 0) a) Study the continuity of f in R. b) Compute the partial derivatives of f at the point 0, 0). c) At which points is f differentiable? a) Note that On the other hand since, do not coincide, the limit 0 lim ft, 0) lim t 0 t 0 t 0 t sin t sin t lim ft, t) lim t 0 t 0 t lim 1 t 0 t lim fx, y) x,y) 0,0) does not exist, and hence f is not continuous at 0, 0). The function is continuous at R \ {0, 0)} since it is a quotient of continuous functions and the denominator does not vanish there. b) The partial derivative with respect to x is since sin0) 0. Similarly, ft, 0) f0, 0) 0, 0) lim t 0 t 0 lim t 0 t 3 0 f0, t) f0, 0) 0 0, 0) lim lim t 0 t t 0 t 3 0 c) First, we note that f is differentiable on R \{0, 0)}, since the partial derivatives exist and are continuous there. The function is not differentiable at 0, 0) because it is not continuous at that point Consider the function fx, y) { x3 y x +y cosxy) if x, y) 0, 0) 0 if x, y) 0, 0) a) Find the partial derivatives of f at the point 0, 0). b) Prove that f is continuous on all of R. Hint: Note that for x, y) 0, 0) we have that 1 x + y 1 x + y c) Is f differentiable at 0, 0)? a) The partial derivatives are and ft, 0) f0, 0) 0 0, 0) lim lim t 0 t t 0 t 3 0 f0, t) f0, 0) 0 0, 0) lim lim t 0 t t 0 t 3 0

4 4 b) The function is continuous on R \ {0, 0)}, since it is a quotient of continuous functions and the denominator does not vanish. Let us study the continuity at the point 0, 0). Let ε > 0. Take δ ε/. If 0 < x + y < δ then, x 3 y x + y cosxy) x x y x + y cosxy) c) The function is differentiable at 0, 0) if x y since x x + y y cosxy) 1 x ) ) y x + y x + y < δ ε fx, y) f0, 0) f0, 0) x, y) lim 0 x,y) 0,0) x + y Since, f0, 0) 0, f0, 0) x, y) 0, the function is differentiable at 0, 0) if lim x,y) 0,0) x 3 y x + y ) x + y cosxy) 0 Given ε > 0, we take δ ε. If 0 < x + y < δ then, x 3 y x + y ) x + y cosxy) x 3 y x + y ) x + y so f is differentiable at 0, 0). x x y x + y ) x + y x y x x + y y x + y y x + y < δ ε 3-7. Compute the derivatives of the following functions at the given point p along the vector v a) fx, y) x + xy 3y, p 1, ), v 3, 4). b) gx, y) e xy + y tan 1 x, p 1, 1), v 1, 1). c) hx, y) x + y ) 1/, p 0, ), v 1, 1). a) x + xy 3y ) x1,y 1 + y, x 6y) x1,y, 10). So, the derivative along the vector 3, 4) is, 10) 3, 4) b) e xy + y arctan x) x1,y1 ye xy + 1, e + π 4 ). So, the derivative along the vector 1, 1) is y 1+x, xe xy x1,y1 + arctan x) e + 1, e + arctan 1) e + c) x + y ) 1/ ) x0,y 1, 1) is e + 1, e + π 4 ) 1, 1) 1 π 4 ) x, y 0, 1). So, the derivative along the vector x +y ) x +y ) x0,y 0, 1) 1, 1) Let Bx, y) 10x x 1 xy + y be the profits of a firm. Last year the company sold x 4 units of good 1 and y units of good. This year, the company can change slightly the amounts of the goods x and y it sells. If it wishes to increase its profit as much as possible, what should x y be? 10x x xy + y) x4,y 10 x y, x + ) x4,y 1, 3)

5 Since the gradient points in the direction of maximum growth of the function, if there is an increase x, y), for the function to increase the most, we must have that x, y) k1, 3). From here we obtain that x k y y 3k. Hence, x/ y 1/ Knowing that, 3) 7 and D 1, )f,3) 3, find, 3) and D vf,3) with v 3, 4 ). We know that and also Letting we have that But this is equivalent to And, therefore We may compute now D 1, )f, 3) D 3, 4 ) f, 3) ), 3),, 3) 1,, 3) 7 z, 3) 7, z) 1, ) z, 3) 4 3 ) 3 ) 3, 3),, 3), 4 ) 3 7, 4), 4 ) Find the derivative of fx, y, z) xy + z y, along the vector v 1, 1, ) at the point 1, 1, 0). Determine the direction which maximizes resp. minimizes) the directional derivative at the point 1, 1, 0). What are the largest and smallest values of the directional derivative at that point? The gradient of the function fx, y, z) xy + z y at the point 1, 1, 0) is f1, 1, 0) xy + z y) x1,y1,z0 y, xy + z, zy ) x1,y1,z0 1,, 0) The derivative along v is D v f1, 1, 0) f1, 1, 0) v 1,, 0) 1, 1, ) 1 The direction which maximizes the directional derivative is f1, 1, 0) f1, 1, 0) 1 1,, 0) and the maximum value of the directional derivative is f1, 1, 0). Likewise, the direction which minimizes the directional derivative is f1, 1, 0) f1, 1, 0) 1 1,, 0) and the minimum value of the directional derivative is f1, 1, 0) Consider the function fx, y) x + y + 1 y gx, y) x + y, ay). Determine: a) The value of a for which the function f g grows fastest in the direction of the vector v, 7) at the point p 1, 1). b) The equations of the tangent and normal lines to the curve xy x + y + x 6 at the point 4, ). Consider the functions fx, y) x + y + 1 y gx, y) x + y, ay)

6 6 a) Their composition is fgx, y)) fx + y, ay) x + y) + a y + 1 and the gradient at the point 1, 1) is fg1, 1)) x + y, x + y + a y ) x1,y1 4, 4 + a ) If we want that direction of the vector v, 7) is the direction of maximum growth of fgx, y)) at the point 1, 1), we must have that v and fgx, y))1, 1) are parallel. That is, 4 + a 7 4 whose solution is a ± b) Note first that the point 4, ) satisfies the equation xy x + y + x 6. Now, the gradient of the function gx, y) xy x + y + x 6 at the point 4, ) is Thus, the equation of the tangent line is g4, ) y 4x +, xy + 1) x4 7, 17) y 7, 17) x 4, y ) 0 and the parametric equations of the normal line are xt), yt)) 4, ) + t 7, 17) 3-1. Find the Jacobian matrix of F in the following cases. a) F x, y, z) xyz, x z) b) F x, y) e xy, ln x) c) F x, y, z) sin xyz, xz) a) The Jacobian matrix of F is b) The Jacobian matrix of c) The Jacobian matrix of D F x, y, z) D F x, y, z) ) yz xz xy xz 0 x ) ye xy xe D F x, y, z) xy 1/x 0 yz sin xyz xz sin xyz xy sin xyz z 0 x ) Using the chain rule compute the derivatives r in the following cases. a) z x xy + y, x r + θ, y r θ b) z x y, x r cos θ, y r sin θ θ a) r θ r + r x y x + y 0 θ + θ x y x + y) 4x y) 8θ

7 7 b) r θ r + r 10x x y cos θ 10y x y sin θ r cos θ r r sin θ r r r θ + θ 10x x y + 10y x y 10x x y r sin θ) 10y r cos θ) x y r cos θ sin θ r r cos θ sin θ r Using the capital K at time t generates an instant profit of Bt) 1 + t) 1/ K Suppose that capital evolves in time according to the equation Kt) 10e t/4. Determine the rate of change of B. Since we see that dk dt 30et/4 d dt B 1 + t) 1/ K t) 1/ dk dt t) 1/ e t/ t) 1/ e t/ Verify the chain rule for the function h x y + y z + z x with x et, y e t and z e t Verify the chain rule for the composition f c in the following cases. a) fx, y) xy, ct) e t, cos t). b) fx, y) e xy, ct) 3t, t 3 ). a) The functions are fx, y) xy and ct) xt), yt)) e t, cos t). Therefore, fxt), yt)) fe t, cos t) e t cos t and d dt fxt), yt)) et cos t e t sin t Now, we compute fct)) dc dt On the one hand, fx, y) y, x) and Therefore, dc dt et, sin t) fct)) dc dt yt)et xt) sin t which coincides with the computation above.

8 8 b) The functions are fx, y) e xy and ct) xt), yt)) 3t, t 3 ). Therefore, fxt), yt)) f3t, t 3 ) e 3t and d dt fxt), yt)) 1t4 e 3t Now, we compute fct)) dc dt On the one hand, fx, y) ye xy, xe xy ) and Therefore, dc dt 6t, 3t ) fct)) dc dt 6ye xy t + 3xe xy t ) x3t 1t 4 e 3t,yt Write the chain rule h x) in the following cases. a) hx) fx, ux, a)), where a R is a parameter. b) hx) fx, ux), vx)). a) b) h x) h x) x, ux), vx)) x, ux, a)) + + x, ux, a)) ux, a) x, ux), vx)) u x) + x, ux), vx)) v x) Determine the points at which the tangent plane to the surface z e x 1) +y is horizontal. Determine the equation of the tangent plane at those points. Consider the function of 3 variables gx, y, z) e x 1) +y z We are asked to compute the tangent plane to the level surface A {x, y, z) R 3 : gx, y, z) 0} at the point x, y, z) where this tangent plane is horizontal. At that point we must have that Since, gx, y, z) gx, y, z) 0, 0, 1) we must have that x 1, y 0. The z coordinate is ) x 1)e x 1) +y, ye x 1) +y, 1 z e x 1) +y x1,y0 1 And the tangent is horizontal at the point 1, 0, 1). The equation of the tangent plane is z Consider the function fx, y) xe y ) 3. a) Compute the equation of the tangent plane to the graph of fx, y) at the point, 0). b) Using the equation of the tangent plane, find an approximation to 1, 999e 0,00 ) 3.

9 9 a) We are asked to compute the tangent plane to the graph of f at the point, 0, f, 0)), 0, 8). Consider the function of 3 variables The graph of f is the level surface of g, gx, y, z) x 3 e 3y z A {x, y, z) R 3 : gx, y, z) 0} Therefore, it is enough to compute the tangent plane to the level surface A at the point, 0, 8). Since, g, 0, 8) 3x e y, 3x 3 e y, 1) x,y0,z8 1, 4, 1) the equation of the tangent plane is that is, 1, 4, 1) x, y, z 8) 0 1x + 4y z 16 b) We are asked to estimate the function fx, y) x 3 e 3y at the point 1 999, 0 00). Since, f is differentiable and that point is very close to, 0), we use Taylor s first order approximation around the point, 0), and we obtain that z 1x + 4y 16 f1 999, 0 00) 1x + 4y 16) x1 999,y Compute the tangent plane and normal line to the following level surfaces. a) x + xy + y z 0 at the point 1, 1, ). b) x + y z 0 at the point 1,, ). c) y x )y x ) z 0 at the point 1, 3, ). a) We compute the gradient x + xy + y z) x,y,z)1,1,) x + y, x + 4y, 1) x,y,z)1,1,) 4, 6, 1) Thus, the equation of the tangent plane is that is, b) We compute the gradient 4, 6, 1) x 1, y 1, z ) 0 4x + 6y z x + y z) x,y,z)1,,) x, y, 1) x,y,z)1,,), 4, 1) Thus, the equation of the tangent plane is that is, c) We compute the gradient, 4, 1) x 1, y, z ) 0 x + 4y z y x )y x ) z) x,y,z)1,3,) xy x ) 4xy x ), y 3x, 1) x,y,z)1,3,) Thus, the equation of the tangent plane is that is, 10, 3, 1) 10, 3, 1) x 1, y 3, z ) 0 10x 3y + z Let f, g : R R be two functions with continuous partial derivatives on R. a) Show that if g x, y) x, y) at every point x, y) R, then f g depends only on y.

10 10 b) Show that if g x, y) x, y) at every point x, y) R, then f g depends only on x. c) Show that if f g)x, y) 0, 0) at every point x, y) R, then f g is constant on R. d) Find a function f : R R such that x, y) yx + x + y, x, y) y x + y, f0, 0) 1 Are there any other functions satisfying those equations? The functions f and g are of class C 1. a) Let a, b), x, b) R. Let hx, y) fx, y) gx, y). By the Mean Value Theorem, hx, b) ha, b) hc) x a, 0) for some point c c 1, c ) tx, b) + 1 t)a, b) tx + 1 t)a, b) with 0 < t < 1. Since, h c) 0 we have that hx, b) ha, b) 0. That is, hx, b) ha, b) for every x R and the function h does not depend on y b) Is very similar to the previous case. c) At each point of R we have that f g) f g) so f g does not depend neither on x nor on y. d) We know that x, y) yx + x + y. Integrating with respect to y, fx, y) yx + x + y)dy 1 y x + xy + y + Cx) where Cx) is a function that depends only on x. The other condition is x, y) y x + y. We try this with the function that we have obtained, ) 1 y x + xy + y + Cx) y x + y + C x) so, C x) 0 and Cx) c, a constant. To find c we use the condition f0, 0) 1. Thus, 0 fx, y) 1 y x + xy + y + c f0, 0) c and f0, 0) 1. Hence c 1 The function fx, y) 1 y x + xy + y + 1 satisfies the above conditions. If there were another function g of class C 1 satisfying the same conditions, we would have that f g)x, y) 0, 0) at every point x, y) R. By part c) there is a constant A R such that f g)x, y) A for every x, y) R. But, since f0, 0) 1 g0, 0), we have that A 0 and the functions coincide.