2t t dt.. So the distance is (t2 +6) 3/2

Transcription

1 Math 8, Solutions to Review for the Final Exam Question : The distance is 5 t t + dt To work that out, integrate by parts with u t +, so that t dt du The integral is t t + dt u du u 3/ (t +) 3/ So the distance is (t +) 3/ 3/ 3/ 3/ 5 3 (33/ 7 3/ ) Question : By the fundamental theorem of calculus (in combination with the chain rule), the derivative is dy dx + (x) 3 When x, this gives a slope of for the tangent line When x, we have y, and so the tangent line is the line of slope passing through (, ), which is y x Question 3: a) Use integration by parts Take u x, so that dv sin(x) Then v cos(x), and du dx The formula gives x sin(x) dx x cos(x) cos(x) sin x + cos(x) dx x + + C b) Integration by parts again Use u ln(x) so dv x 5 dx Then v x and du dx x The formula gives us x 5 ln(x) dx x ln(x) x x dx x ln(x) x 3 dx For the definite integral, ( x x 5 ln(x) ln(x) dx c) This one needs partial fractions Write ) x ln() x A x + B x + A(x + ) + B(x ) Plugging in x, we find B / Plugging in x, we get A / The integral then becomes: 9 x dx 9 ( / x / ) x + dx ((ln 8 ln ) (ln ln 3)) d) Substitute u x +, so that du x dx Then x x + dx u du ln u + C ln(x + ) + C (Notice that x + is always positive, so we don t have to worry about taking the absolute value) e) Rewrite cos 3 x as ( sin x) cos x Then substitute u sin x Answer: 3

2 Question : (a) The integral converges It s given by L dx L x + L tan (x) (b) This one is L L tan (L) tan () π π π L L 3 dx x 3 ln x L L The it is infinite, which means that the integral diverges Question 5: We have dy + x dx dx, and we want to compute x 3 ln L L + ( dy dx) dx, which is: Question First convert to feet: 8 inches is /3 of a foot We know that F kx, and since at 8 inches the force is, the spring constant is k /(/3) 5 Then the work in stretching it to foot from is 5x dx 5 Question 7: We need to integrate b ρga(y)d(y) dy The bounds are 5 to, since we are a trying to put water into the top of the tank The density is ρ A(y) 5π is the area of a cross-section, and doesn t depend on y At last, D(y) y is just the distance the water has to move from height to height y As a result, the work is 5 ()(98)(5π)y dy 9875π (the units here are newtons) Question 8: Since x, the total rainfall is about ( ) (units are inches) Question 9: a) The area is given by write it out The final answer is ( x (x x)) dx This is not such a hard integral, so I won t b) Using the method of washers, the volume is given by the integral π (( x+) (x x + ) ) dx (We had to use a + on each integral, since we re rotating it around the line y, which adds to the radii in question) Question : a) Absolutely convergent by ratio test Taking the absolute values, we find a k+ a k 3k+ k! (k + )! 3 3 k k +

3 Then a k+ k a k, and the ratio test tells us that it converges absolutely b) Absolutely convergent Again we throw out the signs to test absolute convergence; regular k k old convergence then follows In this case, we have + k < We know that k k3/ converges by the p-test, and so this series converges by the direct comparison test k 3/ c) First let s check for absolute convergence; we want to know whether k+ converges By it comparison, we know that this sum and k k k either both converge or both a diverge, since k k b k k k k+, which isn t But we also know that k diverges, by the p-test with p / So the series in question does not converge absolutely On the other hand, it is conditionally convergent by alternating series test: the terms are decreasing and have it, which in the case of an alternating series is everything we need to know d) Use it comparison test to compare with the divergent series that either both series converge or both series diverge We know again that k k ; as before, we see k k diverges, according to the p-test, and so the series diverges e) Divergent by the divergence test: the it of the terms here is not, and so the series must diverge Question : ( ) n 5 n+ n n n a) 5 n 5 n 5 n 5 n /5 k k k k k b) Rewrite this as (n + 3)(n + ) n +, using a partial fractions decomposition n + 3 We recognize this as a telescopic series: ( S k ) ( ) ( + + k + ) k + 3 k + 3 Then k S k, which is the sum of the series c) The simplest way to do it is to note that e k k /e < Thus this geometric series converges to e Question : We know well the series for this function: it s k x + x + x + x 3 + x + is a geometric series with ratio ek This is geometric, so it converges if the ratio x is less than, which means that the interval of convergence is (, ) Differentiating both sides, we find that ( x) + x + 3x + x 3 +

4 We want negative that, so our answer is going to be ( x) x 3x x 3 Question : a) Radius of convergence: Interval of convergence: [, ) To get this, use the ratio test: a n+ n a n (x + ) n+ n n n (n + ) n+ (x + ) n n + n n (x + ) x + This is less than if < x < But we still need to test the endpoints At x, the series is n ( )n This is the alternating harmonic series that we have come to n know and love, and it converges If x, the series is, which is the (not alternating) n harmonic series and diverges The the interval is [, ) b) Radius of convergence: Interval of convergence: [3, 3 + ] We use the usual strategy on this one a n+ n a n n+ (x 3) n+ n n (n + ) n (x 3) n (n + ) n (x 3) n x 3 This is less than if 5 < x < 35 We still need to check the endpoints, as usual At 5, it s ( ) n, while at 35 it s The latter converges by the p-test, so the former does n n as well (in fact it converges absolutely) So the interval of convergence is [5/, 7/] Question : First rewrite x 3 x + x3 ( x/) Now, using the power series representation we get the power series representation x k x k ( x/) ( ) k x k k ( ) k k x k Finally, the desired power series is obtained by multiplying through by x3 : x 3 x + k ( ) k k+ xk+3 The radius of convergence is and the interval of convergence is (, ) Question 5: Use the known power series e x x k k! k

5 a) Substitute x into the power series for e x and get e x k ( ) k xk k! b) Integrate on both sides above to get e x dx K + k Question : The Taylor polynomial of degree about x is ( ) k x k+ (k + )k! P (x) f() + f ()(x ) + f () (x ) + x Substituting x 5 into the polynomial we get the approximation (x ) Question 7: The tangent line has slope y (t) It goes through the point (3, ) when t We have x (t) x (t) 3t + t and y (t) t At t these are 7 and, so the slope is /7 The line with slope /7 going through the point (3, ) is y /7(x 3), which is therefore the slope of the tangent line Question 8: Remember that a plain old unit circle is x(t) cos t, y(t) sin t We to rescale the whole thing by a factor of and then to move the center to the right and up by 3 Our equations are x(t) + cos t x(t) 3 + sin t Question 9: (a) Here s a plot courtesy of the computer: 8 - -

6 (b) We have sin θ sin θ cos θ, so this is r cos θ cos θ sin θ, or r sin θ Multiplying by r we get r r sin θ, so that x + y y, which is the equation for a circle (c) The slope of the tangent line is given by f (θ) sin θ + f(θ) cos θ f (θ) cos θ f(θ) sin θ For us, f(θ) + 3 cos θ Plugging in θ π, we get infinity, because the denominator is that s telling us that the tangent is vertical (d) The tangent line is vertical when f (θ) cos θ f(θ) sin θ, and horizontal when f (θ) sin θ + f(θ) cos θ It is possible to figure out when this happens, but I wouldn t worry too much about it it s a big mess of trig identities (e) Half of one leaf goes from θ to θ π/, so the area is π/ A cos (5t) dt π Question : [ ] [ ] Let A be the matrix Let B be the matrix Let x be the vector 3 For each of the following, either compute it or explain why it isn t defined (a) [ ] [ ] AB 3 The dimensions don t match up right, so this isn t defined (b) [ ] Ax 3 (c) [ ] 5 [ ] [ ] Bx, 3 another one that isn t defined (d) A + B is also not defined, since A and B aren t the same size Question : (a) The augmented matrix is [ ] (b) We add times the first row to the second and get [ ] This is upper-triangular, so we re done The second equation now says that y, so y / The first says x + y, so x and x

7 (c) In matrix form, the equation is [ ] [ ] x y [ ] (d) The inverse matrix is [ ], using our formula (remember that the in the fraction is just the determinant) (e) To solve the system, we use [ ] x [ ] [ ] [ ], y / which matches our answer from the row-reduction method Question : (a) We have [ ] [ ] [ ] 3 3 Ax, 3 3 [ ] [ ] 3 Ay 3 (b) It stretches a vector by a factor of 3 in both directions, and then reflects it over the x-axis (c) This is a diagonal matrix (like the ones we dealt with in lecture), so the eigenvectors are [ ] with λ 3 and [ ] with λ 3 [ 3 ]