Math 103, Review Problems for the First Midterm

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1 Math 0, Review Problems for the First Mierm Ivan Matić. Draw the curve r (t) = cost, sin t, sint and find the tangent line to the curve at t = 0. Find the normal vector to the curve at t = 0.. Find the equation for the set of points p which are equidistant from the plane z = and the point (0, 0, ). Sketch and describe the surface consisting of the points p.. Sketch the curve given by r = sin(θ) and write down the integral representing the length of one loop of the curve. Do not evaluate the integral.. Calculate the it if it exists: x y x 6 + y Evaluate the its, or show that they don t exist: (a) (b) (c) (d) (e) (f) (x,y,z) (0,0,0) (x,y,z) (0,0,0) xy x + y ; xy x + y ; xy x + y ; x + y + z x + y + z (xy+yz+zx); xsin y x + y ; x (y + z ) x + y + z. 6. Assume that α is a real number between 0 and. (a) If x, y > 0, prove the inequality αx+( α)y x α y α. (Hints: Apply ln to the both sides of the inequality and use the fact that ln x is concave.) (b) What inequality do you get if you set α = /? If you in addition put x = a and y = b, what is the name of the obtained inequality? (c) What inequality do you get if you set α = /, x = a and y = b 8? (d) Calculate x sin y x + y Find the curvature of the curve with parametric equations: t ( ) t ( ) x = sin 0 πθ dθ, y = cos 0 πθ dθ. 8. Find the length of the polar curve r = e θ, 0 θ π. 9. Suppose the three coordinate planes are all mirrored and light ray given by the vector a = a, a, a first strikes the xz-plane, and after that zy, and finally the xy plane. Prove that the resulting ray is parallel to the initial ray. 0. Suppose that a 0. (a) If a b = a c, does it follow that b = c? (b) If a b = a c does it follow that b = c?

2 (c) If a b = a c and a b = a c does it follow that b = c?. Find the distance between the lines x = +t, y = +6t, z = t and x = +s, y = 5+5s, z = +6s.. Prove that d ( r (t) [ r (t) r (t)] ) = r (t) [ r (t) r (t)].. Determine the set of points at which the function f(x, y) = xy x +xy+y, if (x, y) (0, 0), 0, if (x, y) = (0, 0) is continuous.. If u = e a x +a x + +a n x n, where a + + a n =, prove that u x + u x + + u x = u. n 5. Let f(x, y) = x y xy x +y, if (x, y) (0, 0), 0, if (x, y) = (0, 0). Prove that f xy (0, 0) = and f yx (0, 0) =. xy, if (x, y) (0, 0), 6. Let f(x, y) = x +y 0, if (x, y) = (0, 0). Prove that f x (0, 0) and f y (0, 0) both exist but the function is not differentiable at (0, 0). 7. Suppose that the equation F(x, y, z)=0 implicitly defines each of the three variables in terms of the other two: x = f(y, z), y=g(z, x), z = h(x, y). If F is differentiable and F x, F y, F z are all nonzero, prove that z x x y y z =.

3 Hints and Solutions. The velocity vector is r (t) = sint, sint cost, cost. The velocity vector at t = 0 is 0, 0,. The equation of the line is x =, y = 0, z = t. The acceleration vector is a (t) = r (t) = cost, cos t sin t, sint. We will need a (0) =,, 0. The speed v(t) can be calculated as v(t) = r (t) = +sin t cos t. Therefore κ(0) = r (0) r (0) / = 5. We have that dv = sint cost(cos t sin t) +sin t cos t and dv (0) = 0. From the equation a (0) = dv (0) T (0)+κ(0) v (0) N (0) we get N (0) = 5,, 0. Problem Problem Problem. Let p(x, y, z) be the equation of our point. Its distance from the point (0, 0, ) is equal to x + y +(z+) and its distance from the plane z = is equal to z. Hence the equation is z = x + y +(z+). After simplifying we get z = x + y. That is elliptic paraboloid.. First we draw the picture as shown above, and then we see that the parametrization of one loop of the curve is The length is equal to 5π L = π x(t) = ( sin(θ))cosθ y(t) = ( sin(θ))sinθ π θ < π. ( cos(θ)cosθ ( sin(θ)sinθ) +( cos(θ)sinθ +( sinθ)cosθ) dθ.. The it is 0. Use the substitution x = r cosθ, y = r sinθ. Then (x, y) (0, 0) if and only if r 0. We have: x y r cos θ sin θ = x 6 + y 6 r 0 r cos 6 θ + sin 6 θ. We now use the fact that cos 6 θ + sin 6 θ = cos 6 θ + ( cos θ) = cos θ + cos θ = cos θ( cos θ) = ( ) cos θ sin θ. Notice that cos θ sin θ = (cosθ sinθ) cos θ+sin θ = hence cos6 θ + sin 6 θ =. We now have and by the squeeze theorem we may conclude that 0 r cos θ sin θ r cos 6 θ + sin 6 θ r r cos θ sin θ r 0 r cos 6 θ + sin 6 θ = 0

4 5. (a) The it doesn t exist since the its along the paths x = 0 and y = x are different. (b) The it doesn t exist since the its along the paths x = 0 and x = y are different. (c) The it is 0. Hint: Use the Squezze Theorem and the inequality x + y xy. (d) The it is 0. (squezze and x + y + z xy + yz + zx, the last inequality is easily obtained by adding x + y xy, y + z yz and z + x zx.) siny (e) The it is 0 use = and x + y xy. y 0 y (f) x +y +z = (x /+y )+(x /+z ) (x y +x z ) = x (y +z ). Now we can easily apply squeeze theorem. This problem can also be solved passing to spherical coordinates. 6. (a) Since lnx is concave function (review calculus ), then ln(αx +( α)y) α lnx+( α)ln y. This is equivalent to the required inequality. (b) For α = / we get x+y xy which is the inequality between the arithmetic and quadratic mean. (c) We get a + b8 a b. (d) We will use that t 0 sint t = 0. Therefore x sin y x + y 8 = y = x y ( siny x + y 8 y ( ) ( x ) ) 8 y x + y 8 Notice that ( ) ( y x 0 x + y 8 ) 8 y y (x + y 8) x + y 8. The squezze theorem now implies that the required it is r (t) = sin ( πθ ), cos ( πθ ). Now use the formula for the curvature. 8. Using the formulas find r (t), r (t), and v(t) = r (t). Then evaluate π 0 v(t). 9. After striking the xz plane the new ray will be given by the vector v = a, a, a. After hitting the yz plane the vector becomes v = a, a, a, and after the xy plane it becomes v = a, a, a. Obviously the vectors a, a, a and a, a, a are parallel. 0. (a) No. Counter-example: a =,, b = 0, 0, c =,. (b) No. Counter-example: a =,,, b =,,, c =,, (c) Yes. Proof: From the first equality we have a ( b c ) = 0 and from the second we have a ( b c ) = 0. If θ is the angle between a and b c we conclude that a b c sinθ = 0 and a b c cosθ = 0 Since we assumed that a 0 we can cancel these equalities by a. After cancelling we square both of them and add together to use the fact that cos θ + sin θ =. We get: b c = 0 hance b = c.. The vector of the first line is n =, 6, and of the second is n =, 5, 6. These two vectors are not parallel. It is easy to check that the two lines don t intersect. The distance between the two lines is equal to the distance between two planes α and β defined in the following way: α is the plane containing the first line and parallel to the second; β is the plane containing the second line and parallel to the first. A normal vector for both planes is n n. Once we find this vector we have the equation of the two planes. Then pick any point from α and find its distance from β.

5 . We use the product rule for the dot and cross product. Then d ( r (t) [ r (t) r (t)] ) = r (t) [ r (t) r (t)]+ + r (t) [ r (t) r (t)]+ r (t) [ r (t) r (t)]. The quantity r (t) [ r (t) r (t)] represents the volume of the parallelepiped determined by the vectors r (t), r (t), and r (t), which is 0 since two of the vectors are 0. Similarly, the second quantity is 0 and the required result immediately follows.. The function is obviously continuous at all points (x, y) (0, 0). At the point (0, 0) we can see that doesn t exist, hence the function is not continuous at (0, 0).. We immediately find u x = a e a x + +a n x n,, hence xy x + xy+y u x n = a n e a x + +a n x n. Similarly u = a x ea x + +a n x n,, u = a x ne a x + +a n x n n u x + + u x = (a + + a n) e a x + +a n x n = u. n 5. For (x, y) (0, 0) we can easily calculate f x (x, y) using the rules of differentiation. We get f x (x, y) = y x +x y y (x +y ). Similarly, for (x, y) (0, 0) we get f y (x, y) = x y +x y x (x +y ). However f x (0, 0) and f y (0, 0) can t be calculated this way. We have to use the definition of the partial derivatives: f((0, 0)+(h, 0)) f(0,0) f(h, 0) 0 f x (0, 0) = = h 0 h 0 h = +0 = 0. Similarly f y (0, 0) = 0. Hence We now want to calculate f xy (0, 0): f x (x, y) = f y (x, y) = y x +x y y (x +y ), (x, y) (0, 0) 0, (x, y) = (0, 0), x y +x y x (x +y ), (x, y) (0, 0) 0, (x, y) = (0, 0). f x (0, h) f x (0, 0) h 0+0 h h 0 (0 f xy (0, 0)(0, 0) = = +h ) h 5 = 5 =. Similarly we get f yx (0, 0) =. f (h,0) f (0,0) 6. Using the definition we find that f x (0, 0) = = 0 and similarly f y (0, 0) = 0. However, the function is not differentiable because the directional derivative of f in the direction of the vector u =, is ( ) f h, h f(0, 0) D u f(0, 0)= = = DNE. 7. Using the formulas z x = F x F z, x y = F y F x, y z = F z F y. Multiplying these togethere gives z x x y y z =. 5