A small review, Second Midterm, Calculus 3, Prof. Montero 3450: , Fall 2008

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1 A small review, Second Midterm, Calculus, Prof. Montero 45:-4, Fall 8 Maxima and minima Let us recall first, that for a function f(x, y), the gradient is the vector ( f)(x, y) = ( ) f f (x, y); (x, y). x y A critical point of the function f is a point (x, y ) such that ( f)(x, y ) = (, ), that is, a point for which Let us also define the quantity We have the following D(x, y) =. Theorem. Assume ( f)(x, y ) = (, ). f x (x, y ) = and f y (x, y ) =. ( ) ( ) ( ) f f f (x, y) (x, y) (x, y). x y x y a) If D(x, y ) > and then (x, y ) is a local minimum. b) If D(x, y ) > and then (x, y ) is a local maximum. f x (x, y ) > f x (x, y ) < c) If D(x, y ) < then (x, y ) is a saddle point. d) If D(x, y ) = it is not possible to determine the nature of the critical point. Example. Find all the critical points of the function f(x, y) = x 4 + 8x + 48xy y, and determine the nature of each of them.

2 Solution: We compute and To find the critical points we set from where we get the equations f x (x, y) = x + 4x + 48y = (x x 4y), f (x, y) = 48x 4y = 4(x y). y f f (x, y) = and (x, y) =, x y x x 4y = and x y =. From the second equation we obtain y = x. We plug this into the second equation to obtain x x 8x = x(x x 8) =. The solutions of this equation are x =, or x = ± 4 +, so we get x = 4 or x =. This gives us the points (, ), (4, 8) and (, 4). Next we compute the second derivatives of the function f. We obtain f x (x, y) = (x 4x), f (x, y) = 4 and y f (x, y) = ( 4) = 48. x y This gives us For (, ) we have D(x, y) = () (x 4x) (48) = 44(6x 8x 6). D(, ) = 44 ( 6) < so (, ) is a saddle point. For (4, 8) we have D(4, 8) = 44(96 6) > and f (4, 8) = (48 6) <. x

3 We conclude that (4, 8) is a local maximum. Finally, D(, 4) = 44( ) > and f (, 4) = 4( 6 4) <. x (, 4) is also a local maximum. lagrange multipliers. To find the extreme values of a function f(x, y) subject to the condition v(x, y) =, we introduce an additional variable λ, and solve the system ( f)(x, y) = λ( v)(x, y) and v(x, y) =. Example. Find the maximum area of a rectangular box inscribed in an equilateral triangle of sides of length a (see Figure ), that sits symmetrically relative to one height of the triangle. Figure : Box inside an equilateral triangle, symmetric relative to one height. Solution: The left and right vertices of the triangle have coordinates ( a, ) and ( a, ) respectively, so the upper vertex has coordinates ( a, ) (the triangle is equilateral). This means that the side of the triangle in the first quadrant is part of the line with equation y = x + a.

4 Say the upper right corner of the rectangle has coordinates (x, y). Then the area of the rectangle is f(x, y) = xy. In this case, the upper right corner of the rectangle must be on the line y = x + a. so our constraint is We then compute and v(x, y) = y + x a =. ( f)(x, y) = (y, x), ( v)(x, y) = ( ; ). We set ( f)(x, y) = λ( v)(x, y) which gives us the equations y = λ and x = λ. This obviously means that y = x. We plug this into the equation y + x a = to obtain x = a, which gives us x = a and y = a. The maximum area of such rectangle is then 4 4 A = a 8, or half the area of the triangle. Absolute max, min Recall that a region D in the plane is bounded if it is contained in a large enough disk. On the other hand D is closed if it contains its boundary. We have the. Theorem. If the function f is continuous over the region D, and D is closed and bounded, then f attains both its global max and its global min in D. To find the global max, min of a function f over a given region we follow these steps:

5 a) Find the critical points of f in D by finding f and setting it equal to (, ). b) Find the extreme points of f on the boundary of D. c) Compare the values of f at the points found in the two previous points. That with the largest value is where the global max is located, and the one with the smallest value is where the global min occurs. Example. Find the global max and min of f(x, y) = x + y 4x y over the region D = {(x, y) : x and y x}. Solution: Let is first get an idea of what the region D is. To do this we notice that D is bounded below by the line y =, by the line x = on the right, and by the line y = x above. In other words D is a triangle (see figure ). Line y = x Point (,) Figure : The region D from Example is the shaded area.

6 Now we need to find the critical points of the function f. To do this we compute ( f)(x, y) = (4x 4, y ). Obviously the only point for which ( f)(x, y) = (, ) is (, ). Next, we need to find the extreme values of f on the boundary of D. This boundary consists of three portions: the lines y =, x = and y x =. To find the max/min of f over these lines we can use Lagrange multipliers, one line at a time. For the like y = we take v(x, y) = y = as our constraint. The system we ought to solve here is ( f)(x, y) = λ( v)(x, y) along with the condition y =. We find that ( v)(x, y) = (, ), so we obtain the equations 4x 4 = λ = and y = λ. This means x =. Since we have also the equation y =, the point we find here is (, ), which is actually part of the lower boundary of the triangle D. Next, we look at the like x =. Here we can use the constraint v(x, y) = x =. We find ( v)(x, y) = (, ), so we get the equations 4x 4 = λ and y =. This gives us y =, and we still have x =, so we get the point (, ). We next consider the line = x. Here we can take the constraint v(x, y) = y x =. Then ( v)(x, y) = (, ) so we obtain the equations 4x 4 = λ, and y = λ. These equations together give us that λ = x and also that λ = y. This means that x+y = 4, or x+y =. But we still have y = x, so we can solve this system to obtain x = and y = 4. We obtain the point (, 4 ), which is located on the slanted edge of the triangle D. Finally, we also have the vertices themselves, which we ought to consider. We

7 compute f(, ) = 5 f(, ) = 4 f(, ) = f(, 4 ) = 4 9 f(, ) = f(, ) = and f(, 4) = 6. The global min is located at (, ) and the global max occurs at (, 4). Example 4. Find the global max/min of f(x, y) = x + 4y + x y over the region D = {(x, y) : x 4 + y }. Solution: First we find ( f)(x, y) = (x +, 8y ). The only solution to ( f)(x, y) = (, ) is (, 4 ). Next, we analyze the boundary of D. Here the boundary is defined by the equation x 4 + y =, so we can take We obtain v(x, y) = x 4 + y. ( v)(x, y) = ( x, y ). The system ( f)(x, y) = λ( v)(x, y) gives us the equations x + = λx and 8y = λy. We multiply the first equation by, divide the second by and rearrange both a bit to obtain (4 λ)x = and (4 λ)y =. This means that both x and y. Dividing the first equation by x and the second by y

8 we obtain 4 λ = x = y. This means y = x. We use this in the equation x 4 + y = to obtain y =, so y = ±. Since y = x, we obtain the points (, ), and (, ). We finally compute f(, 4 ) =, f(, ) = 4 and f(, ) = 4 +. Obviously (, ) is where the max occurs, and the min is at (, 4 ). Integrals of functions of two variables Let us first notice that the region D = [a, b] [c, d] is a rectangle. For the function f(x, y) defined on D, we have D f da = b a ( d c ) f(x, y) dy dx = d c ( b a ) f(x, y) dx dy. Furthermore, if the function f(x, y) can be written as a product f(x, y) = u(x)v(y) of a function u(x) (that depends only on x) and a function v(y) (that depends only on y), then D ( b f da = a ) ( d ) u(x)dx v(y) dy. c Example 5. Let D = [, ] [, ], and Find D f da. f(x, y) = + x + y. Solution: Obviously f(x, y) = u(x)v(y) if we define u(x) = + x and v(y) = + y.

9 Hence We now compute D ( ) ( ) f da = ( + x dy )dx. + y ( + x )dx = ) (x + x = 8 and We conclude that dy + y = arctan(x) = π 4. D f da = 8 π 4 = π. We can integrate over more general regions. For example we can consider Ω = {(x, y) : a x b and ψ (x) y ψ (x)}. In this case we obtain If we had instead, we get f da = b Ω a ψ (x) ( ) ψ (x) f(x, y) dy dx. Ω = {(x, y) : c y d and ξ (y) x ξ (y)}. f da = d Ω c ξ (y) ( ) ξ (y) f(x, y) dx dy. The value f da represents the volume of the region below the graph of f, and above the Ω region Ω. We also have area(ω) = da. Example 6. Evaluate the integral below by changing the order of integration: Ω ( ) y + sin dy dx. x Solution: The region of integration in this case is Ω = {(x, y) : x and x y }. The right edge of this region (see figure ) corresponds to the curve y = x, which can also be

10 written as x = y. In fact, from the picture we note that the region can also be described as From here we integrate x and this last integral we compute as y Ω = {(x, y) : y and x y }. ( ) y + sin dy dx = ( ) y + sin dx dy = = y ( ) y + sin dx dy, ( y ( ) ) y + sin dx dy ( ) y + sin y dt ) = ( y cos + = ( ( ) ) cos cos (). Figure : Region of integration for Example 6. Polar coordinates We consider here different variables. We set x = r cos(θ) and y = r sin(θ). This implies x + y = r. We change dxdy by rdrdθ.

11 Example 7. Integrate the function sin( x + y ) over the annular region Ω = {(x, y) : x + y 4}. Solution: The main issue here is to express the region of integration in polar coordinates. Clearly Ω = {(r, θ) : r and θ π}. We then integrate sin( x + y ) dxdy = π Ω = π sin(r)r dθdr sin(r) r dr. We can find this last integral by parts, using for instance u = r and dv = sin(r) dr, so that du = dr and v = cos(r). This yields sin(r)r dr = r cos(r) + cos(r) dr = cos() cos() + sin() sin(). We then conclude that sin( x + y ) dxdy = π(cos() cos() + sin() sin()). Ω Example 8. Evaluate the volume of the solid bounded above by z = x + y, inside the curve x + y = ax. Here a > is a positive constant. Solution: Using polar coordinates x = r cos(θ) and y = r sin(θ), so that x + y = r, we notice that the curve x + y = ax can be described as r = a r cos(θ), or r = a cos(θ). We need to know now what are the appropriate limits of integration. By completing the square in x + y = ax, we obtain (x a) + y = a, so this curve is in fact the circle of radius a, centered at (a, ) (see Figure 4). From here we obtain the following region of integration: Ω = { π θ π and r a cos(θ)}.

12 Since z = x + y, we obtain z = x + y = r, so V = = π π π π = 8a a cos(θ) 8a r dr dθ cos (θ) dθ π π cos (θ) dθ. To find the antiderivative cos (θ) dθ, we integrate by parts using u = cos (θ) and dv = cos(θ) dθ. Then du = cos(θ) sin(θ) dθ and v = sin(θ). This gives us cos (θ) dθ = cos (θ) sin(θ) + cos(θ) sin (θ) dθ = cos (θ) sin(θ) + cos(θ)( cos (θ)) dθ = cos (θ) sin(θ) + cos(θ) dθ cos (θ) dθ = cos (θ) sin(θ) + sin(θ) cos (θ) dθ. From here we obtain cos (θ) dθ = cos (θ) sin(θ) + sin(θ) cos (θ) dθ, so cos (θ) dθ = cos (θ) sin(θ) + sin(θ), and we obtain π π cos (θ) dθ = 4, so the volume we seek to compute is V = a 9. Example 9. Find the volume of the solid bounded by the cylinder x +y = b, and the planes z = and y + z = a. Here we assume a > b >. Solution: We first note that, because a > b >, the plane y + z = a hits the plane z =

13 (a,) Figure 4: Region of integration for Example 8. outside the cylinder x + y = b. This means that the region of integration, in polar coordinates, is given by Ω = { r b and θ π}. The function we need to integrate is z = a y, which in polar coordinates can be expressed as f(r, θ) = a r sin(θ). Then V = = b π b π (a r sin(θ))r dθ dr a r dθ dr b π = πab. r sin(θ) r dθ dr