Midterm Exam II Math 1320 - Engineering Calculus II April 6, 2015 Answer each question completely in the area below. Show all work and explain your reasoning. If the work is at all ambiguous, it is considered incorrect. No phones, calculators, or notes are allowed. Anyone found violating these rules will be asked to leave immediately. Point values are in the square to the left of the question. If there are any other issues, please ask the instructor. By signing below, you are acknowledging that you have read and agree to the above paragraph, as well as agree to abide University Honor Code: Name: Signature: uid: Solutions Question Points Score 1 15 2 15 3 20 4 20 5 20 6 10 Total: 100 Note: There are 6 questions total on the exam, as well as a challenging bonus question at the end of the exam that is worth trying if and only if you have extra time.
1. Consider the following two vectors: u = h3, 2i, v = h2, 0i. 5 (a) Find an expression for the angle between u and v and draw graphical representation of these vectors on a standard (x,y) coordinate system. Solution: We know that this is just an immediate application of the dot product. Specifically, u v = ku kvk cos, where is the desired angle between the two vectors, meaning that u v = cos 1, kukkvk thus we just need to compute each of these. Our dot product can be obtained by taking the sum of the products of the components: u v = h3, 2i h2, 0i =3 2+2 0=6. Also, the magnitudes can be computed from the distance formula: p kuk = 3 2 +2 2 = p p 13, kvk = 2 2 +0 2 =2. Thus, our is: u v = cos 1 kukkvk = cos 1 6 2 p 13. u v 10 (b) Compute proj v u and draw the graphic representation of this operation and its result. Solution: Recall that proj v u means the projection of the vector u onto the vector v. That is, we want to compute the component of u that is in the direction of v. This corresponds to the following picture: The length of the vector we want, denoted comp v u is the following: comp v u = u v kvk = 6 2 =3. Thus, we want a vector of length comp v u in the direction of v, so we just compute the unit vector in the direction of v: v kvk = 1 h2, 0i = h1, 0i. 2 Therefore, our total projection is: proj v u = comp v u v = h3, 0i. kvk 2/11 /15 pts
Notice, this makes sense. Since v is just in the x direction, the projection of u onto v is just the x component of u. u v proj v u 3/11 /0 pts
2. Consider the two following points: P 1 =(1, 2, 3) P 2 =( 1, 0, 4). 5 (a) Find a vector representation of the line that passes through both P 1, P 2. Solution: We first need to find the vector that points in the direction of the line through P 1 and P 2, which is just the displacement vector between these two points:! P 1 P 2 = hp 2 P 1 i = h 2, 2, 1i. Note, this is our v for our line. The equation of a line is then where r 0 = hp 1 i = h1, 2, 3i, sowehave: r = r 0 + vt, r = hx,y,zi = h1, 2, 3i + h 2, 2, 1it = h1 2t,2 2t,3+ti. 10 (b) Consider the following vector: v = h1, 1, 1i. Find the equation of the plane that contains the line described in part (a) that also is in the direction of the vector v. Solution: We want a plane that is in the direction of the line from the previous problem, which I ll call v 0 = h 2, 2, 1i and now also v = h1, 1, 1i. We need the normal vector to the plane, which is orthogonal to both of these and therefore can be obtained via the cross product: n = v 0 v = i j k 2 2 1 1 1 1 = h 3, 3, 0i. For a plane, we need a point and a normal vector, so we can just choose a point on the line, say P 1 =(1, 2, 3), which gives us that our plane equation is: which is the same as:! n PP 1 =0, h 3, 3, 0i hx 1,y 2,z 3i = 3(x 1) + 3(y 2) + 0(z 3) = 0. 4/11 /15 pts
3. Match the following equations to one of the surfaces on the next page. If you explain/show work, you may get extra credit even for a wrong answer. Note I : not all of the equations are given in rectangular coordinates. Note II: not all of the surfaces have corresponding equations, but every equation has a corresponding surface. 5 (a) x 2 y 2 + z 2 =1. 5 (b) z = r. 5 (c) x = z 2 + y 2. 5 (d) cos =0. Solution: The easiest way to identify this one is that notice due to the negative signs, you obtain hyperbolas. Specifically, when z =0, no solutions exist. This is clearly (v). Solution: We ve talked many times about this example in cylindrical coordinates. It s the classical cone, (iii). Solution: Again, taking slices can yield the answer, but notice that it s effectively something that grows in the x-direction, the only of which, is (viii). Solution: This is probably the trickiest of the bunch, but if you recall your coordinate changes, we know that z = cos, so this is just z =0, the plane described by (i). 5/11 /20 pts
Midterm Exam 2 Math 1320 - Engineering Calculus II (i) (v) (ii) (vi) (iii) (vii) (iv) (viii) 6/11 April 6, 2015 /0 pts
4. Consider two bees flying around in a beehive, where the first bee s position at time t is described by r 1 (t) =ht, t,1i, and the second bee s trajectory is r 2 (t) =ht 2,t 3, 5+ti. 10 (a) Colliding would be catastrophic for both of the bees. Do they ever have to worry about this? That is, do the two bees ever collide? If so, at what time? Solution: To check if the trajectories intersect, we simply have to see if the x,y,z components are all the same at any given time. Thus, if we first examine the x component, we check t = t 2 =) t 2 t =0, which only occurs at t =0and t =1, two possible candidates. Now, if we check the y component, which, for the bees to collide, requires t = t 3 =) 2t = 3, which only occurs at t =3/2, we see that the bees never collide. We do not even need to check the z component. 10 (b) Suppose the hive is a sphere of radius 3, centered at the origin, which we can describe by x 2 + y 2 + z 2 =3 2. Does the first bee ever escape the sphere? That is, does r 1 ever intersect the sphere? If so, when? Solution: Similar to a lab question, to see if the first bee intersects the sphere, we simply check if the position of the bee r 1 ever satisfies the equation of the sphere. To do so, we plug in each of the x,y,z coordinates for x,y,z in the sphere equation: which after simplifying, yields (t) 2 +( t) 2 +1 2 =3 2, 2t 2 +1=9 =) t = ±2. Clearly, we do not care about t = 2, so the bee leaves the hive at t =2. 7/11 /20 pts
5. Consider the following vector equation: 5 (a) Compute T(t), the unit tangent equation. r(t) =3ti +2t 2 j. Solution: We ve defined T to be the following: T(t) = r0 (t) kr 0 (t)k, so we clearly just need to take the derivative to find r 0 (t), which is, in this case Computing the magnitude of this, we see r 0 (t) =h3, 4ti. kr 0 (t)k = p 3 2 +(4t) 2 = meaning that we have our unit tangent vector T(t) = p 9 + 16t 2, r0 (t) kr 0 (t)k = 1 h3, 4ti. 9 + 16t2 5 (b) Compute N(t), the unit normal equation. Solution: This calculation is a little messier, but at the end of the day, still just basic calculus. We ve defined the unit normal to be N(t) = T(t) kt 0 (t)k, so we need to take our answer from the previous question and differentiate it. Note, for the y component, you do have to use the quotient rule, but what we find is T 0 48t (t) = (9 + 16t 2 ) 3/2, 36 12 (9 + 16t 2 ) 3/2 = (9 + 16t 2 h 4t,3i. ) 3/2 Therefore, we can now compute the magnitude noticing that we can pull the scalar out in front out, to see that we get kt 0 (t)k = 12 p ( 4t) (9 + 16t 2 ) 3/2 2 + (3) 2 = 12(9 + 16t2 ) 1/2 (9 + 16t 2 ) 3/2 = Therefore, we now have our unit normal vector 12 9 + 16t 2. N(t) = T(t) kt 0 (t)k = 1 p h 9 + 16t 2 4t,3i. 8/11 /10 pts
10 (c) Compute the curvature, apple(t). Solution: For this problem, you could feasibly use the definition apple(t) = kt0 (t)k kr 0 (t)k, since we already have both of these, but we can also use the other definition, which is an easy calculation in this case. The other definition we ve used is apple(t) = kr0 (t) r 00 (t)k kr 0 (t)k 3. We know what r 0 (t) is, so we can take another derivative to find that Thus, we now have the following: r 00 (t) =h0, 4i. Thus, our final answer is simply r 0 (t) r 00 (t) = i j k 3 4t 0 0 4 0 = h0, 0, 12. apple(t) = kr0 (t) r 00 (t)k kr 0 (t)k 3 = 12 p 9 + 16t 2 3 = 12 (9 + 16t 2 ) 3/2. 9/11 /10 pts
10 6. In class, we proved that if r(t) = c, wherec is a constant, then r(t)? r 0 (t) for all t. Prove the converse. That is, prove the following statement: If r 0 (t)? r(t) (that is, the vectors are orthogonal) for all t, then r(t) = c, a constant. For this problem, it s sufficient to consider r(t) to have the form r(t) =hf (t),g(t),h(t)i. Hint: What does it mean for two vectors to be orthogonal? Write this down and then integrate. Solution: Let s consider the form provided in the problem, that is r(t) =hf (t),g(t),h(t)i, then r 0 (t) =hf 0 (t),g 0 (t),h 0 (t)i. For two vectors to be orthogonal, this means that their dot product is zero, in other words: 0=r(t) r 0 (t) =f (t)f 0 (t)+g(t)g 0 (t)+h(t)h 0 (t). On the left side if integrate, we just get a constant, c: Z 0 dt = c. The right side of the equation is screaming to be integrated, yielding, by the chain rule: Z Z Z f (t)f 0 (t) dt + g(t)g 0 (t) dt + h(t)h 0 (t) dt = f (t)2 + g(t)2 + h(t)2 2 2 2. Thus, we re left with: 2c = f (t) 2 + g(t) 2 + h(t) 2, which means that p 2c = p f (t) 2 + g(t) 2 + h(t) 2, but what is this object on the right hand side? It s exactly the magnitude of r(t), that is: kr(t)k = p f (t) 2 + g(t) 2 + h(t) 2, so we just have that some constant p 2c, which we can now call a different constant C: p 2c = C = kr(t)k, which is exactly what we hoped to prove. 10/11 /10 pts
7. (Extra credit) Prove that a circle of radius a has constant curvature. Solution: This just Example 3 in section 10.3 of the book, found on page 710. 11/11 /0 pts