Practice problems. 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b.

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1 Practice problems 1. Given a = 3i 2j and b = 2i + j. Write c = i + j in terms of a and b. 1, 1 = c 1 3, 2 + c 2 2, 1. Solve c 1, c Suppose a is a vector in the plane. If the component of the a in the direction of 1, 1 is 2 and the component of a in the direction of 2, 1 is 7 5/5. Find a vector with length 3 that points in the opposite direction as a does. Assume a = x, y. Then, using the two components, you can write out two equations for x and y. Find x and y. the vector desired is 3a/ a. 3. Given a = 1, 1 and b = 2, 3, decompose b into two parts, such that one is parallel with a and one is perpendicular with a. b = b a a a a. b = b b = b (b a)a. a 2 4. a, b and c are three vectors in space. (1). If a, b, c are perpendicular with each other and all are unit vectors (one example is {i, j, k}), for an arbitrary vector d, find the numbers C 1, C 2, C 3 in terms of the four vectors: Dot each side with a, b and c d = C 1 a + C 2 b + C 3 c. (2). Answer the same question for general three vectors a, b, c provided det(a, b, c) = (a b) c 0. (Hint: Dot both sides with some suitable vectors.) Dot each side with a b etc. 5. v 1, 2, 3 = 5, 2, 3 and v 2, 1, 2 = 3, 0, 3. Find v. If v is perpendicular with both a and b, then v = λa b. You can plug it back to determine λ. 6. The area of the parallelogram determined by v and w is 5. Both v and w are perpendicular with 1, 2, 2. Compute v 1 v 2 v 3 det w 1 w 2 w

2 (Hint: det(a, b, c) = (a b) c = a (b c) = b (c a)) The question is simply asking (v w) 1, 1, 2. From the conditions, v w = (s)5 1,2,2 1,2,2 where s = 1 or s = 1 depending on which direction it is pointing. the answer is therefore , 2, 2 1, 1, Let A(5, 2, 3), B(6, 4, 0), C(7, 5, 1) and D(14, 14, 18). (1). Compute the area of the triangle ABC. (2). Compute the volume of the parallelepiped determined by AB, AC and AD. (3). Find the distance from D to plane ABC (You can just use the results from (1) and (2)) (1). 1 2 AB AC (2). V = ( AB AC) AD = determinant (3). d = V/ AB AC 8. Given two planes x + y + z = 5 and 3x y = 4, write a parametric equation and a symmetric equation for the line of intersection of the two planes. There are essentially 3 ways to find the intersection of line. Quiz 1. Check 9. Find the line that passes through (2, 6, 3) and parallel with both 3x 9y + 4z = 10 and 2x y + 4z = 12. The direction of the line is parallel with the cross product of the normal vectors of the planes (why?). 10. Consider the skew lines L 1 : x(t) = 3 2t, 4t 1, 3t + 2 and L 2 : x(t) = 6t + 3, 1 t, 2 2t. We have two ways to find the distance tween them: (1). Find a point on L 1 called P and point on L 2 called Q such that P Q is perpendicular with both of them. (Hint: without loss of generality, we can assume P (3 2t 1, 4t 1 1, 3t 1 +2) and Q(6t 2 +3, 1 t 2, 2 2t 2 ). Then,vector P Q should be perpendicular with both.) (2). Find any point P on L 1 and any point Q on L 2. Find a vector that is perpendicular with both lines using cross product, called n. Then, d = P Q n / n. 2

3 Verify that these two methods give the same answer. The problem has given enough hints. 11. (1). Consider the curve r(t) = 3t sin(t), 3t cos(t), 2t 2. Compute the arclength of this curve from t = 1 to t = 2. (2) Suppose that T (t) is the unit tangent vector of a curve. Explain why dt /ds is perpendicular with T where s is the arclength. (3). Compute the curvature of r(t) = e t cos ti + e t sin tj + e t k. (1). s = 2 1 r (t) dt (2). Use the fact that T is a constant. (3). This was shown in lecture. 12. Consider r(t) = t, t 2, t 3. Find the tangent line of this curve at (1, 1, 1) and a plane that is normal to this curve at (1, 1, 1). A vector tangent to the curve is the velocity v(t) = r (t). For the line, the line passes through (1, 1, 1) and is parallel with v(1); for the plane, it passes thorough (1, 1, 1) and is perpendicular with v(1). 13. Parametrize y = x 2 as a vector valued function r(t) and then compute the curvature at x = 0. For function y = f(x), one parametrization is r(x) = x, y = x, f(x). You can rename x to be t. 14. (1). Describe the surface y 2 = x 2 + z 2 and get a rough sketch. (Hint: This is a surface of revolution.) (2). Find the trace of the ellipsoid x 2 /a 2 + y 2 /b 2 + z 2 /c 2 = 1, which is tangent to planes P 1 : x = 2, P 2 : y = 3 and P 3 : z = 1, in the plane that passes through (0, 0, 0) and is perpendicular with (1, 1, 0). (1). This is the cone revoluted from y = ±x, z = 0 about y-axis. (2). a = 2, b = 3, c = 1. The plane is simply x y = 0. hence, the trace is x2 4 + x2 9 + z2 1 = 1, y = x. This is an ellipse. *************************************************************** 1. Sketch the level curves of f(x, y) = x 2 y 2. The level curves are x 2 y 2 = k. Discuss k > 0, k = 0, k < 0 3

4 2. Can we define a suitable value for the function f(x, y) at (0, 0) to make it continuous? If yes, compute the value; if not, explain why. (a). f(x, y) = xy x 2 +y 2 (b). f(x, y) = arctan( 1 x 2 +2y 2 ) (c). f(x, y) = x2 +y 6 y 2 +x 6 The key idea is to check if the limits exist as (x, y) (0, 0). If the limits exist, we can define the values to be the limits and then the functions become continuous. Part(a). The limit is zero as we showed in lecture. You can use polar coordinates. (b). The inside goes to positive infinity but the arctan then goes to π/2. The third one, by checking different directions y = kx. You see different limits. 3. Consider the function f(x, y) = (1). Compute f x (0, 0) and f y (0, 0). { sin(x 2 + y 2 ) x > 0, y > 0 0 otherwise (2). Is the function continuous at (0, 0)? (3). (Bonus) Is the function differentiable at (0, 0)? f(h,0) f(0,0) (1). Use the definition f x = lim h 0 h. (2). This is simply asking if f(x, y) f(0, 0) as you approach the origin. 4. Let P = ln x + y 2 and Q = 2xy + y 1 y 2. Can you find a function f(x, y) such that f x = P and f y = Q for x > 0, y < 1? If yes, find f; if not, explain why. This problem tests f xy = f yx. You should check that P y, Q x are continuous. They verify that if they are equal or not. 5. Suppose f(x, y) = ln(x 2 + y 2 ). (1). Compute the tangent plane of the graph at (1, 1, ln 2). Can you find a normal vector of this plane? (2). Compute the tangent line of the level set f(x, y) = ln(2). Can you find a normal vector of this line? (3). On the level set f(x, y) = ln(2), near the point (1, 1), y can be regarded as a function of x. Compute y (x) x=1. 4

5 (1). The tangent plane of the graph is given by z = f(1, 1) + f x (1, 1)(x 1) + f y (1, 1)(y 1). Another way is to regard it as the level set of F = f(x, y) z = 0. Use the technique for computing tangent planes of level sets to compute the tangent plane. (2). The tangent line can be computed using two understandings. The first is the linear approximation: { } f(1, 1)+f x (1, 1)(x 1)+f y (1, 1)(y 1) ln(2) = 0 f x (1, 1)(x 1)+f y (1, 1)(y 1) = 0. The second understanding is that f(1, 1) is a normal vector of the line. Then, f(1, 1) x 1, y 1 = 0 (3). This tests implicit differentiation. y (1) = fx(1,1) f y(1,1) 6. Suppose w = 2u 2 + 3v and u = s + t and v = 2s 3t. Compute w s t at s = 1, t = 2 in two ways. This tests chain rule. It s straightforward. 7. Let f(x, y, z) = ln(1 + x 2 + y 2 z 2 ). (1). Find the direction in which f(x, y, z) decreases the fastest at (1, 1, 1). (2). Compute the rate of change of f with respect to distance in the direction indicated by 2, 1, 4 at (1, 1, 1). (3). Let r(t) = t 2, t 2, 2t 2 1. Compute r(1) and r (1). Then, compute d dt f(r(t)) t=1 using what you got just now. Explain how this is related to the directional derivative in (2). (1). As we explained, f points the fastest increasing direction while f points the fastest decreasing direction. In our case, we need a unit normal vector. Hence, the unit vector has something to do with f. (2). This is just directional derivative: D u f, but you should find u first. (3). Chain rule: d dt f = f(r(1)) r (1). You only need to compute f,r(1) and r (1). This rate of change w.r.t. time is equal to the directional derivative times speed. The speed is v = r (1). If you multiply this with the directional derivative, you get the same thing. 5

6 8. (1). Suppose P V nrt = 0 where n and R are two constants. Show that P V T V T P = 1 (2). Let F (x, y, z) = ln(x 2 + y 2 ) e x z. Consider the level set F = ln 2 1. This determines an implicit function. Near point (1, 1, 1), compute x y. This problem tests implicit differentiation. Let F = P V nrt. Then, we have a level set F (P, V, T ) = 0. The product is 1. P V = F V F P V T = F T F V T P = F P. F T You can verify it directly: P = nrt/v. Hence, P/ V = nrt/v 2. Similarly you can compute others. The product is 1. (2). Use the formula for implicit differentiation. 6