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2xy . x2 + y 2 Solution. As f is given by quotient of two polynomials functions, we can evaluate the polynomials for any (x, y) ∈ R2 . However, if the denominator x2 + y 2 ̸= 0, one can evaluate f (x, y), and hence the domain of f = { (x, y) | x2 + y 2 ̸= 0} = R2 \ {(0, 0)}, i.e. the set of all points in R2 except (0, 0). (ii) For any point (x, y) ∈ D, if either one of x and y is zero, then f (x, y) = 0. Suppose that m = xy where x and y are not zero, then we have the following 2mx2 2m f (x, y) = f (x, mx) = 2 = . In order to determine the range R, 2 2 x +m x 1 + m2 we apply the discriminant method as follows: for any value u in the range, one 2m has u = 1+m 2 for some real number m ∈ R, then one knows that the quadratic equation um2 − 2m + u = 0 in m has at least one real root, so its discriminant ∆ = (−2)2 − 4(u)(u) = 4(1 − u2 ) ≥ 0. Hence one has −1 ≤ u ≤ 1. So the range of f is the closed interval [−1, 1], Remark. It would be easier to use the polar coordinates (x, y) = (r cos θ, r sin θ). 1. Determine (i) the domain and (ii) the range of the function f (x, y) = 2. Sketch the k-level curve the function of g(x, y) = 9x2 − 4y 2 if k = −9, −4, 0, 4, 9. Solution. Recall that the level curve Lc of g(x, y) at c is given by the set of points (x, y) which takes g-value equal to c, i.e. Lc = { (x, y) | g(x, y) = c }. (x, y) lying on or outside the ellipse 4x2 + 9y 2 = 36. (ii) The domain of g(x, y, z) consists of all points in 3-dimensional space R3 except the origin (0, 0, 0), i.e. Dom(f ) = R2 \ {(0, 0, 0)}. 6. Draw the graph of (i) f (x, y) = x and (ii) g(x, y) = y 2 . Solution. Recall that the graph of z = f (x, y) is the set of points in R3 given by (x, y, f (x, y)), where (x, y) is the input and the third coordinate f (x, y) is the corresponding output. (i) (ii) 7. Consider the line ℓ : x = 3 − t, y = −1 + 2t, z = 5 + 8t. Find the equations of the projection of the line ℓ onto (i) the xy-plane. Solution. (i) Let P (x, y, z) be any point in the space, after projecting it onto the xy-plane, its image Q(x, y, 0). So the image ℓ0 of the line ℓ is given by ℓ0 : x = 3 − t, y = −1 + 2t, z = 0. 3. √ Describe (i) the domain and (ii) the graph of the function f (x, y) = −1 + 16 − x2 − y 2 . Solution. (i) The domain D of f (x, y) is given by 16 − x2 − y 2 ≥ 0, i.e. D is the closed circular disc with center at (0, 0) with radius 4. (ii) The graph of f is a set { (x, y,√ f (x, y) ) | (x, y) ∈ D }, so if (x, y, z) in the graph of f, one has z = f (x, y) = −1 + 16 − x2 − y 2 , i.e. (z + 1)2 = 16 − (x2 + y 2 ), i.e. the graph of f consists of the points in the upper hemisphere x2 +y 2 +(z−1)2 = 42 with center at (0, 0, 1) and radius 4. 8. Consider the line ℓ : x = 3 − t, y = −1 + 2t, z = 5 + 10t. (i) Determine the point P where the line ℓ meets the plane S : x − y + 3z + 8 = 0. (ii) Find the equation(s) of the projection of the line ℓ onto the plane S : x − y + 3z + 8 = 0. Solution. (i) First we ﬁnd the point P0 of intersection of the line ℓ and the plane S as follows: 0 = (3 − t) − (−1 + 2t) + 3(5 + 10t) + 8 = 27t + 27 i.e. 27t = −27 and t = −1. The intersection point is P (x, y, z) = (2, −3, −5). (ii) The image of the projection of the line ℓ onto S means that any point Q on ℓ will projected to S along the normal direction n of S, hence the image R lies on S, and the vector −−→ QR is parallel to n. 4. Let f (x, y) = (x2 + y 2 )1/3 be a function. Find out the domain and the range of the function f (x, y). Solution. For any point (x, y) in xy-plane, i.e. (x, y) ∈ R2 , we have x2 + y 2 ≥ 0, √ √ 3 3 2 2 2 so x + y is a real number and x + y 2 ≥ 0, and hence f (x, y) ≥ 0. From this, we know domain of f is the entire xy-plane R2 . And we expect the range of f is equal to the set of all non-negative real numbers, i.e. any non-negative number ℓ (ℓ ≥ 0) can be expressed as the value√of f evaluated at some point (x, y) √ 3 in R2 . In fact, f (ℓ3/2 , 0) = 3 (ℓ3/2 )2 + 02 = ℓ3 = ℓ. √ 5. Determine the domain of (i) f (x, y) = 4x2 + 9y 2 − 36, and g(x, y, z) = √ 2xyz2 2 . x +y +z √ Solution. (i) The domain of f (x, y) = 4x2 + 9y 2 − 36 consists of all the points When t = 0, we have a point Q(3, −1, 5) on ℓ, but Q is not on the plane S. It follows from the equation of S that n = (1, −1, 3) is normal to S. Let ℓ′ be a line through Q with direction vector n, then the equation of ℓ′ is given by −−→ OQ + tn = (3 + t, −1 − t, 5 + 3t). Suppose ℓ0 meets the plane S at a point R, then 0 = (3 + t) − (−1 − t) + 3(5 + 3t) + 8 = 11t + 27, i.e. t = 27/11, and so R(3+27/11, −1−27/11, 5+3×27/11) = (60/11, −38/11, 136/11). As both points P (2, −3, −5) and R(60/11, −38/11, 136/11) are on S, then the ℓ′ is the line P R, y+3 x−2 z+5 Then the equation of the line ℓ0 : is given by 60/11−2 = −38/11+3 = 136/11+5 . The following are taken from an old test paper. 1 x2 9. Let u(x, t) = √ exp(− ). The domain of u is 4t t A. the entire xy-plane R2 B. { (x, t) | x ∈ R, and t > 0 } C. { (x, t) ∈ R2 | t ̸= 0 } D. { (x, t) | x > 0, and t > 0 } E. None of above Solution. (x + y)2 is x2 + y 2 B. R2 \ {(0, 0)} C. x2 + y 2 ≤ 1 10. The domain of the function f (x, y) = A. the entire xy-plane R2 E. None of the above 4xy is x2 + y 2 A. [−1, 1] B. [−2, 2] C. all positive real numbers E. None of the above. D. x2 + y 2 ≥ 1 11. The range of f (x, y) = D. all negative real numbers