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Course: MATH 153, Spring 2008
School: Ohio State
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Word Count: 1230

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FINAL MATH PRACTICE 153 SP01 1. Prove (using formal definition of convergence) the following assertion: n lim 3 = 1 n Proof: take solve) &gt; 0; to have the limit equal to 1 we have to check (read: n | 3 - 1| &lt; But n 3 &gt; 1, so we can drop the absolute value thing: n 3-1&lt; n 3&lt;1+ Take the natural log, and get 1 ln(3) &lt; ln(1 + ) n ln(3) &lt;n ln(1 + ) Since we get n...

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FINAL MATH PRACTICE 153 SP01 1. Prove (using formal definition of convergence) the following assertion: n lim 3 = 1 n Proof: take solve) > 0; to have the limit equal to 1 we have to check (read: n | 3 - 1| < But n 3 > 1, so we can drop the absolute value thing: n 3-1< n 3<1+ Take the natural log, and get 1 ln(3) < ln(1 + ) n ln(3) <n ln(1 + ) Since we get n bigger than a number ... we're all set. The limit IS 1. Date: 05/31/2000. 1 2 MATH 153 SP01 2. Check whether the following series is convergent or not (point out what test you are using!) 1 n(ln(n))2 Proof: since it involves ln it's a good chance integral test might work. First of all, notice that the general term DECREASES, and converges to 0. So we can use the integral test. 1 dn n(ln(n))2 2 1 Use t = ln(n) dt = n dn and so the integral becomes: 1 t-1 1 1 dt = =- =- 2 t -1 t ln(n) Plug in and get 0 ... so the integral is convergent, so the series is convergent. PRACTICE FINAL 3 3. Find the radius of convergence and the interval of convergence for the following power series nxn Proof: use root test. n lim n n=1 so radius is R = 1 = 1. 1 To find the interval of convergence we need to check 1 (the endpoints of (-1, 1)). For both cases note that the general term converges to infinity (for 1) or diverges (for -1). In any case, it does NOT converge to 0, hence the series cannot converge. So, the interval of convergence is (-1, 1). 4 MATH 153 SP01 4. Choose a good parametrization and set up the formula for the length of the following curve x3 = y 2 where -3 y 2. 2 3 Proof: Use y = t and so x = 3 y 2 = t2 = t 3 . To find the length use the formula: 2 length = -3 2 -3 (x )2 + (y )2 dt 2 1 ( t- 3 )2 + 12 dt 3 PRACTICE FINAL 5 5. Transform into polar coordinates (if necessary) and set up the formula for computing the area between the following two curves: x2 y 2 + =1 4 9 and r = 1 + cos() (note: assume they do not intersect!) Proof: x = r cos(t) and y = r sin(t); we get r2 cos2 (t) r2 sin2 (t) + =1 4 9 1 r2 = cos2 (t) sin2 (t) + 9 4 Since we assume they don't intersect (which they probably don't, actually 2 ...) to find the area between them just use the consecrated formula: 0 1 r2 2 2 0 1 1 ( 2 cos2 (t) + 4 sin2 (t) 9 - (1 + cos()) dt We assumed that the ellipse is enclosing the cardioid ... but if it doesn't that's still ok ... take the ABSOLUTE VALUE of the above integral ... 6 MATH 153 SP01 6. Find what conic the following equation represents; point out all pertinent information for this conic AFTER you shift it in standard position (foci, directrix, asymptotes etc) 2x2 - 20x + 8y 2 + 8y = 12 Proof: complete the square(s). (2x2 - 20x) + (8y 2 + 8y) = 12 1 2(x2 - 10x + 25) - 50 + 8(y 2 + y + ) - 2 = 12 4 1 2 2 2(x - 5) + 8(y + ) = 64 2 (x - 5)2 (y + 1 )2 2 + =1 32 8 bring it in standard position x2 y 2 + =1 32 8 It's an ellipse; it has vertices: two are 32 and the other two are 8;two foci: 32 - 8 = 24. PRACTICE FINAL 7 7. Take the following two vectors: i + j and i + j + k. Add to each of them a vector of the form -i - j + ck, where c is a constant. What value should c have so that the resulting vectors are perpendicular? Proof: If you add -i - j + ck to i + j you get ck. If you add -i - j + ck to i + j + k you get (c + 1)k. These two vectors are perpendicular if and only if their dot product is 0 AND IF NONE OF THEM IS ZERO. But if you compute the dot product of those two vectors you get c(c + 1) and if you set it equal to 0 you get: c(c + 1) = 0 c = 0 or c = -1 The first yields case the first vector being zero, the second gives the second is equal to zero. Hence NO value if c works. Actually you can see what happens if you see that the two vectors considered are actually on the same line always: ck and (c+1)k are multiples of k. So they can not be perpendicular. 8 MATH 153 SP01 8. Find the area of the triangle defined by the following three points: (1, 2, 1), (0, 1, 0), (2, 2, 2). Proof: To find the area of a triangle take half the length of the cross product of two vectors that define it (take two of its sides). In our case, since we have three points given, we need to consider "differences" of points to get vectors. (1, 2, 1) - (0, 1, 0) i + j + k (2, 2, 2) - (0, 1, 0) 2i + j + 2k Take their cross product: i j k 1 1 1 2 1 2 = (2 - 1)i - (2 - 2)j + (1 - 2)k = i - k Hence 1 1 Area = ||i - k|| = 2 2 2 PRACTICE FINAL 9 9. Find the equation of the plane passing through the points given in the above problem. Find the distance from the origin to this plane. Proof: Since you produced i - k as the cross product of the two vectors in the plane it means that i - k is a normal vector. Hence, choosing (0, 1, 0) as the point needed (a plane needs a normal vector and a point), we get: x-z =0-0=0 The first thing you should check when finding the distance from a point to a plane is whether the point isn't already IN the plane (in which case the distance is zero) ... let's check whether the origin is in our plane: 0-0=0 which is correct! hence the origin is a point in our plane, so the distance is 0. 10 MATH 153 SP01 10. Let F and G be two vector valued functions. Let H be the vector function defined as follows: H(t) is perpendicular at all times on both F (t) and G(t) and its length is one. Find the derivative of this vector in terms of F and G and their derivatives. Proof: (we worked it out in class ... check your notes) PRACTICE FINAL 11 11. Given an object whose position at all times is given by r(t) = t2 i + sin(t)j + et k find the velocity, speed and acceleration at t = 0. Does the object ever stop accelerating? Proof: velocity = r (t) = (t2 ) i ...

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Ohio State - MATH - 153
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Ohio State - MATH - 153
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Ohio State - MATH - 153
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