PRACTICE PROBLEMS FOR MIDTERM I

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1 Problem. Find the limits or explain why they do not exist (i) lim x,y 0 x +y 6 x 6 +y ; (ii) lim x,y,z 0 x 6 +y 6 +z 6 x +y +z. (iii) lim x,y 0 sin(x +y ) x +y Problem. PRACTICE PROBLEMS FOR MIDTERM I Consider the function f(x, y) = (x ) + (y ). (i) Draw the level diagram for f. (ii) Draw the graph of f. (iii) Using the ɛ δ definition, prove that f is continuous at the origin. Problem 3. Consider the function f(x, y) = x y 4. (i) Carefully draw the level contours (level curves) of f. (ii) Single out the level curve passing through (, ) and draw the gradient of the function at (, ). (iii) Find the equation of the tangent line to the level curve through (, ) at the point (, ). (iv) Compute the directional derivative of f at (, ) in the direction u = ( 4 5, 5) 3. Use this calculation to estimate f((, ) +.0u). (v) Find the unit direction v of steepest descent for the function f at (, ). decrease in that direction? (vi) Find the two unit directions w for which the derivative D w f(, ) Problem 4. Consider the function f(x, y) = ln(e x y 3 ). What is the rate of (i) Find the partial derivatives of the function f at (, ). Write down the equation of the tangent plane to the graph of f. (ii) Using part (i), find the approximate value of the number ln(e 4. (.0) 3 ). Problem 5. Give an example of a continuous function f : R R and a compact set K R such that the preimage of K is not compact. Problem 6. Let f : R R be a function which admits a continuous derivative f. Show that f is uniformly continuous over any interval [a, b]. Problem 7. Show that for all x 0. e x + x + x Problem 8. Let f : R R be a function such that f(x) f(y) x y.

2 Show that f is constant. Hint: Divide by x y and make x y. What is the derivative of f? Problem 9. Let f : R R be a twice differentiable function such that Show that there exists c R such that f (c) =. f(0) = 0, f( ) = f() =. Hint: Apply the mean value theorem to the function g(x) = f(x) x. Problem 0. Show that the function f(x) = sin(x 3 )/x is Lipschitz hence uniformly continuous over (0, ). Problem. Let a,..., a n be real numbers. Show that the set U of k k matrices A such that is invertible is open. (A a I)(A a I)... (A a n I). Solutions (i) The limit does not exist. Indeed, we first calculate the limit when x = y. The limit equals x + x 6 lim x 0 x 6 + x =. Now make x = 0 and y. We have y 6 lim y 0 y = lim y 0 y4 The answers are different, so the limit does not exist. (ii) We have x 6 + y 6 + z 6 x + y + z = x4 x x + y + z + y4 y x + y + z + z4 z x + y + z In each of the three summands above, we have a product of a function that goes to 0 (such as x 4 ) and a fraction which is a bounded function (the bound of each fraction is ). Each limit equals 0 by a theorem proved in class. Therefore, x6 +y 6 +z 6 x +y +z (iii) We have sin(x + y ) x + y sin(x + y ) x + y = for z = x + y. The last limit is computed as follows sin z sin z lim = lim z. z 0 z z 0 z sin z z The function sin(z)/z is bounded by as shown in Homework. The function sin z z z is product of a bounded function by the function z which converges to 0, hence sin z lim z 0 z The original limit has to be smaller or equal in absolute value, hence it also equals 0.

3 . (i) The level curves are (x ) + (y ) = c which are circles of radius c centered at (, ). (ii) The graph is a paraboloid with vertex at (,, 0). (iii) Note first that f(0, 0) =. We pick ɛ > 0. We need δ such that This in turn becomes (x, y) (0, 0) < δ = f(x, y) f(0, 0) < ɛ. (x ) + (y ) < ɛ (x x + ) + (y y + ) < ɛ x(x ) + y(y ) < ɛ. We will pick δ. In this fashion (x, y) < δ = x + y < δ = x < δ, y < δ = x <, y < = x < 3, y < Thus x(x ) + y(y ) x x + y y < δ 3 + δ 3 = 6δ. If we pick δ ɛ/6, we are done. Thus a good choice for δ is δ = min(, ɛ 6 ). (i) The level curves with level c are x y 4 = c = x = cy 4 = x = ± cy. Only levels c 0 are in fact possible. The cases c 0 and c = 0 give different shapes for the level curves. The level curves are two parabolas for c 0 and the line x = 0 for c In fact, one should also remove the origin (0, 0) from all the level curves because the function is not defined there. (ii) The level is f(, ) =. The level curve is f(x, y) = f(, ) = = x = y 4 = x = ±y. The level curve is a union of two parabolas through the origin. The gradient ( ) x f = y 4, 4x y 5 = f(, ) = (, 4) is normal to the parabolas. (iii) The slope of the gradinet is 4 =. The tangent line is perpendicular to the gradient hence it has slope. The equation of the tangent line is y + x = = (y + ) = (x ) = x + y =. (iv) We compute ( 4 D u f = f u = (, 4) 5, 3 ) = 4. 5 For the approximation, we have f(, ) = and f((, ) +.0u) f(, ) +.0f u = =.04. (v) The direction of steepest decrease is opposite to the gradient. We need to divide by the length to get a unit vector: v = f ( (, 4) = f + 4 =, ). 5 5 The rate of decrease is the directional derivative D v f = f v = (, 4) (, ) = 0 =

4 (vi) Write We have Since w has unit length w = (w, w ). D w f = f w = (, 4) w = w + 4w = 0 = w = w. w + w = = ( w ) + w = = w = ± 5. Therefore ( ) w = ±, (i) Using the chain rule, we compute Similarly, We compute e x e x f x = ln(e x y 3 ) = ln(e x y 3 ) = f x(, ) = = = ln e y e x y 3 e x f y = ln(e x y 3 ) = 3 y ln(e x y 3 ) = f y(, ) = 3 = 3 ln e 4 The tangent plane is (ii) The number we are approximating is f(, ) = ln e 4 = 4 =. z = (x ) (y ) = z = x y + 4. f(.05,.0) = = Let f : R R, f(x) = x +. The preimage of the compact set K = [0, ] is the entire real line R since 0 < f(x) < is true for all x. The preimage is not compact (since R is not bounded). 6. We let f(x) = e x x x. We find f (x) = e x x. We have shown in class that e x + x for x 0, hence f (x) 0 for all x 0. Thus f is increasing over [0, ) hence f(x) f(0) = 0 for x 0. This implies f(x) 0 = e x + x + x. 7. Since f is continuous over the compact interval [a, b], it is bounded. A function with bounded derivative is Lipschitz over the interval [a, b], by a homework problem. Such a function is then uniformly continuous over [a, b] (since we proved in class that Lipschitz implies uniformly continuous). 8. We fix x. For y = x + h we have f(x + h) f(x) h = f(x + h) f(x) h h, after dividing by h. Now make h. Then, f(x + h) f(x) h since the fraction goes to 0 faster than h. The above fraction gives the derivative of f. Hence, f (x) exists and f (x) Therefore, f is constant.

5 9. Let g(x) = f(x) x. Note that g(0) = f(0) 0 = 0 and g( ) = g() = f(±) (±) = By the mean value theorem over the interval (, 0), we can find s (, 0) such that g (s) = g(0) g( ) 0 ( ) Similarly, looking at g over the interval (0, ), we can find t (0, ) such that g (t) = g() g(0) 0 By the mean value theorem applied to g over the interval (s, t) we can find c such that g (c) = g (s) g (t) s t However, g = f hence g (c) = 0 = f (c) =. 0. We check that the derivative of f is bounded. Indeed, f (x) = 3x cos(x 3 ) x sin(x 3 ) x x 4 = 3 cos(x 3 ) sin(x3 ) x 3. This expression is bounded. Indeed, 3 3 cos(x 3 ) 3. By a homework problem sin(x3 ) x 3 [We showed in the homework that sin(y) y for y 0. Set y = x 3. We similarly show sin(y) y which was used above also for y = x 3.] Thus 5 f(x) 5, hence f is bounded. Thus f is Lipschitz by another homework problem, hence uniformly continuous (proved in class).. Consider the function f : Mat k R, f(a) = det(a a I) (A a n I). The function f is continuous since it is a composition of continuous functions. Indeed, the determinant is continuous as shown in class, and matrix operations (multiplication, substraction, etc) are continuous since they are polynomial in the entries of the matrix. Therefore, f is continuous. The set U of the problem is the preimage U = f ((, 0) (0, )). Being the preimage of an open set under a continuous map, U is open.