Review Solutions MAT V (a) If u = 4 x, then du = dx. Hence, substitution implies 1. dx = du = 2 u + C = 2 4 x + C.


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1 Review Solutions MAT V. (a) If u 4 x, then du dx. Hence, substitution implies dx du u + C 4 x + C. 4 x u (b) If u e t + e t, then du (e t e t )dt. Thus, by substitution, we have e t e t dt e t + e t u du ln u + C ln et + e t + C. (c) If u sin(z ), then du z cos(z )dz. Therefore, substitution gives z cos(z ) sin(z ) dz u du u + K sin(z ) + K. (d) If u (ln w) and dv dw, then du ln w w by parts yields (lnw) dw (ln w) w w lnw w dw (lnw) w dw and v w. Hence, integration ln wdw. To compute ln wdw, we use integration by parts a second time. If ũ ln w and dṽ dw, then dũ dw. and ṽ w. Thus w (ln w) dw (ln w)w w dw (ln w)w w + C. w Combining this with our first result gives (ln w) dw (ln w) w w(lnw) + w + C. (e) If u + y, then u + y and u du dy. It follows that dy u du and y u which implies that y + u + u +4. By making a substitution, we have (y + ) u y dy u u u 4 du + 8u du 9 u u 7 + C (+y)5/ (+y) 7 + C. (f) Observe that x + 6x + 5 (x + 6x + 9) + 6 (x + ) + 6. If x + 4u, then dx 4du and (x + ) + 6 6(u + ). By substitution, we have x + 6x + 5 dx arctan(u) 4du + C 6(u ) arctan ( ) x+ 4 + C. page of
2 MAT V Review: page of (g) If u w and dv sin(w) dw, then du w dw and v cos(w). Using integration by parts, we have w sin(w) dw w cos(w) + w cos(w) dw w cos(w) + w cos(w) dw. To compute w cos(w) dw, we use integration by parts once again. If ũ w and dṽ cos(w) dw, then dũ dw and ṽ sin(w), whence w cos(w) dw w sin(w) sin(w) dw w sin(w) + cos(w) dw. 9 Therefore, we have w sin(w) dw w cos(w) w sin(w) + + cos(w) + C. 9 7 (h) If u z z, then du ( z) dz and substitution gives z dz z z u du u + C z z + K. (i) If u ln t and dv t dt, then du dt and v t4. Integration by parts t 4 implies t ln t dt (ln t) t4 t 4 dt t4 lnt t4 + C. 4 4 t 4 6 (j) First, we have y 4 + y + y + y + y + y (y + y + ) + y + y + y + y + y + To find a partial fraction decomposition, we write y + y + A y + + B A(y + ) + B(y + ) y + (y + )(y + ) which implies that A(y + ) + B(y + ) (A + B)y + (A + B). Since the coefficients of the various polynomials must be equal, we obtain A + B and A + B. Solving these two equations gives A and B. Hence, y 4 + y + y + y + y + y + y + y +.
3 Review: page of MAT V Thus y 4 + y + y + dy y + y + y y y + y + y + dy + ln y + ln y + + C + ln y+ y+ + C. (k) If u e t, then du e t dt and substitution gives e t dt e t + e t + du u + u +. To find a partial fraction decomposition, we write u + u + A u + + B u + A(u + ) + B(u + ) (u + )(u + ) which implies that A(u + ) + B(u + ). Setting u implies A and setting u implies B. Thus, du u + u + u + u + du ( ) e t + ln u + ln u + + K ln + K. e t + (l) If u cos θ, then du sin θ dθ and substitution yields sin θ dθ cos θ + cos θ du u + u To find a partial fraction decomposition, we write u + u A u + + B A(u ) + B(u + ) u (u )(u + ) which implies that A(u ) + B(u + ). Setting u implies A / and setting u implies B /. Thus, du du ln u+ u +u u+ u u + C ln cos θ+ cos θ + C. (m) First, observe that x x + x + x x x x x + A x + B x which implies Ax+B(x ). Setting x implies A and setting x implies B. Therefore, we have x x + dx x + dx x x x x x + ln x x + C.
4 MAT V Review: page 4 of (n) To find a partial fraction decomposition, we write s + (s + )(s ) A s + B (s ) + Cs + D s + A(s )(s + ) + B(s + ) + (Cs + D)(s ) (s + )(s ) A(s s + s ) + B(s + ) + (Cs + D)(s s + ) (s + )(s ) (A + C)s + ( A + B C + D)s + (A + C D)s + ( A + B + D) (s + )(s ) Comparing coefficients, we have A + C A + B C + D A + C D A + B + D. The equations A+C and A+C D imply that D and C A. Combining this with the equation A + B C + D, we have A + B. Similarly, the equations A + B + D and D give A + B. From A + B and A + B, we obtain A and B. Therefore, A, B, C, D and the partial fraction decomposition is s + (s + )(s ) s + (s ) + s s + Therefore, we have s + (s + )(s ) ds s + (s ) + s s + s +. s + (s ) + s s + s + ds ln s s + ln(s + ) arctan(s) + C. (o) To find a partial fraction decomposition, we write z +z+4 z +z z +z+4 A + Bz+C A(z +)+z(bz+c) z(z +) z z + z(z +) (A+B)z +Cz+A z(z +). Comparing coefficients, we have A + B, C and A 4 which implies A 4, B and C. Thus, we obtain z +z+4 z +z 4 z + z+ z + 4 z z z + + z +
5 Review: page 5 of MAT V and z + z + 4 z + z dz 4 z z z + + z + dz [ 4 ln z ln z + + arctan(z) ] 4 ln ln 4 + arctan( ) ( ln + arctan()) 4 ln ln 4 + ln + π ln ln + π. (p) If u x, then du dx and x u +. Substitution implies (q) We have w dw w x dx x x + (u + ) du u u + u + u + u u + + u + u du [ u + u + ln u u dw w du ln. w w dw arcsin(w) + w + K. (r) If u y, then du dy and substitution gives / 8y + 4y dy u du + u du + u u du + u [ arctan(u) ln + u ] π 4 ln. (s) Observe that t + 4t (t 4t + ) (t 4t + 4) + (t ). If u t, then du dt and we have dt t + 4t du u arcsin(u) + C arcsin(t ) + C. (t) If u arctan(x) and dv dx, then du dx and v x. Hence, integration +x by parts yields x arctan(x)dx x arctan(x) dx x arctan(x) ln( + +x x ) + K.
6 MAT V Review: page 6 of (u) If u w +, then du dw and substitution gives 4 dw du + (w + ) arctan(u) + C arctan(w + ) + C. + u (v) If u ln z, then du dz. Observe that ln and z ln(eπ/ ) π/. Hence, substitution gives e π/ dz z cos(ln(z)) π/ du cos(u) π/ sec(u) du ln sec(u) + tan(u) π/ ln( + ). (w) If u y, then du dy and du y y dy. Observe that + y + u. Substitution yields 6 dy y( + y) (x) If u (ln t)( + ln t), then du ( + ln t)dt t (ln t)( + ln t) du + u arctan(u) + C arctan( y) + C. (+ln t) dt and substitution gives t du u u + C (ln t)( + ln t) + C. (y) Recall that (7) θ+ e (θ+) ln(7). If u (θ + ) ln(7), then du ln(7)dθ and dθ du. Hence, substitution yields ln(7) (7) θ+ e u eu dθ du ln7 ln7 + C (7)θ+ ln7 + C. (z) If u cos(x), then du sin(x)dx and sin(x)dx du. Using substitution, we obtain sin(x)e cos(x) dx eu du eu + C ecos(x) + C.. (a) If u x n and dv e x dx, then du nx n dx and v e x. Integration by parts implies x n e x dx x n e x n x n e x dx. (b) If u x n and dv cos(ax) dx, then du nx n dx and v sin(ax). Using a integration by parts, we have x n cos(ax) dx a xn sin(ax) n x n sin(ax) dx. a
7 Review: page 7 of MAT V (c) If u cos n (x) and dv cos(x)dx, then du (n ) cos n (x) sin(x) dx and v sin(x). Applying integration by parts, we obtain cos n (x) dx cos n (x) sin(x) sin(x) ( (n ) cos n (x) sin(x) ) dx cos n (x) sin(x) + (n ) sin (x) cos n (x) dx ( cos n (x) sin(x) + (n ) cos (x) ) cos n (x) dx cos n (x) sin(x) + (n ) cos n (x) dx (n ) cos n (x) dx. Therefore, n cos n (x) dx cos n (x) sin(x) + (n ) cos n (x) dx. which implies cos n (x) n cosn (x) sin(x) + n n cos n (x) dx.. (a) If u 5x +, then du 5dx and substitution gives 5x+ dx b 7 du 5u [ ln u 5 ] b 7 ln b ln Therefore, the integral dx diverges. 5x+ (b) If u t and dv e t dt, then du dt and v e t. Using integration by parts, we obtain ( [ te te t dt ] t b Here, we have used the fact that + b [ te t e t] b ) e t dt be b e b +. be b b e b e b and e b. Thus, the integral te t dt converges. (c) By definition, we have dz z + 5 dz z dz z + 5.
8 MAT V Review: page 8 of If z 5u, then dz 5 du and substitution gives b [ ] b dz du arctan(u) arctan(b) π. z +5 5(u +) 5 5 Here, we have used the fact that arctan(b) π. Finally, the substitution w z implies that dz z + 5 dw w + 5 Therefore, dz π. z +5 5 (d) If u ln w, then du dw and substitution gives w dw w ln w b ln dw w + 5. du [ln u u ]b ln ln b ln(ln ). Hence, the integral dw diverges. w ln w (e) If u ln w, then du dw and substitution gives w ln dw w ln w du u b + ln du b u b [ln u ]ln + b Therefore, the integral dw diverges. w lnw (f) If x u, then dx du and substitution gives 4 x dx b b Thus, the integral (g) If u y, then du y π ln(ln ) lnb, b + u du [arcsin(u)]b b arcsin(b) arcsin() π/. b 4 x dx converges to π/. dy and substitution gives π e y dy e u du π y b + e + e b e π. b b + Therefore the integral 4 x dx converges to e π. (h) Observe that is not defined at θ 4. Thus, (4 θ) b 6 dθ (4 θ) a 4 c a dθ (4 θ) + b b dθ (4 θ). If u 4 θ, then du dθ and substitution gives 6 [ ] dθ du b 4 + (4 θ) c + u c + u c + + c. Thus, 6 dθ diverges. (4 θ) c
9 Review: page 9 of MAT V 4. (a) The area of an equilateral triangles with sides of length l is l. Hence, the crosssectional area is A(x) ( sin x) sin x and Volume π A(x)dx π sin x dx [ ] π cos x 4. (b) The area of an square with sides of length l is l. Thus, the crosssectional area is A(x) ( sin x) 4 sin x and Volume π A(x)dx π 4 sin xdx [ 4 cosx] π (a) Since the area of a circle of radius r is πr, the crosssectional area is A(x) π(x ) πx 4 and Volume A(x) dx [ ] πx 4 x dx (b) Since the area of a circle of radius r is πr, the crosssectional area is A(x) π( 9 x ) π(9 x ) and Volume A(x) dx [ π(9 x ) dx π 9x x ] 6π. (c) Since the area of a circle of radius r is πr, the crosssectional area is A(x) π(e x ) πe x and Volume A(x) dx [ ] πe x e dx π x π( e ). 6. (a) Since the function is increasing, lefthand sum will give an underestimate and righthand sum will give an overestimate. (b) Since the function is concavedown, trapezoid rule will give an underestimate and the midpoint rule will give an overestimate. (c) Since the function is concaveup, trapezoid rule will give an overestimate and the midpoint rule will give an underestimate. Furthermore, because the function is decreasing, lefthand sums will also give an overestimate and righthand sums will give an underestimate. (d) Since the function is concaveup, trapezoid rule will give an overestimate and the midpoint rule will give an underestimate.
10 MAT V Review: page of 7. (a) Since y (/)(x + ) /, we have dy x x dx + x 4 + x. Hence, arclength + ( ) dy dx dx + x4 + x dx ( + x ) dx [ + x dx x + x ]. (b) Since x (ey + e y ), we have dx dy (ey e y ) and ( ) arclength + dy dx dy 4 (ey + + e y ) dy e y + e y dy + 4 (ey + e y ) dy (c) Since y x4 +, we have dy 4 8x dx x and 4x arclength + ( dy dx (ey + e y ) dy [ e y e y] e e. ) dy x x 6 dx + ( x [ ] x + x dx 4. 4x 4 8x (d) Since y ln( x ), we have dy dx x arclength / / / + ( ) dy / dx dx + 4x ( x ) dx / ( x + x ) dx / x and 4x ) dx (x + 4x ) dx + ( x x ) dx x 4 +x + ( x ) x + dx x dx / [ x ln x + ln x + ] / ln /. Here, we have used the partial fraction decomposition x + x + x + +x. + x + +x dx
11 Review: page of MAT V 8. (a) The probability that you dropped your glove within one kilometer of home is e x dx [ e x] e. (b) The probability that you dropped it with y kilometers equals to.95 whence.95 y e x dx [ e x] y e y. Solving e y.5 for y, we obtain y ln(.5).5 km. 9. (a) If Q n represents the quantity, in milligrams, of cephalexin in the blood right after the nth tablet, then we have Q 5 (b) In general, we have Q 5(.) Q 5(.) + 5(.) Q n 5(.) n + 5(.) n + + 5(.) [ + (.) + + (.) n ] 5( (.)n ).. (c) To determine the quantity of cephalexin in the body in the long run, we take the it as n tends to infinity: Q 5( (.) n ) n Q n Therefore, the quantity of cephalexin in the body in the long run is 5.5 milligrams.. This is a slightly encrypted geometric series: ( ) n (a) Since ( ) n ( ) n, this is a geometric series. Because <, it converges. n
12 MAT V Review: page of (b) Since + n + n 5 n ( ) n + 5 ( ) n + 5 ( ) n, 5 this is a sum of three convergent geometric series. Hence + n + n converges. 5 n (c) If f(x) xe x then f (x) e x x e x ( x )e x. When x >, f (x) < which implies f(x) is decreasing. Because f(x) xe x is positive continuous and decreasing on [, ), we may apply the integral test: [ xe x dx e x ] b Therefore, the series ne n converges. be b + e e. ( ln(x)+ (d) If f(x) x then f (x) ). When x >, f (x) < which ln x x (ln(x)) / implies f(x) is decreasing. Because f(x) x is positive, continuous and lnx decreasing on [, ), we may apply the integral test: x ln x [ dx ] (lnx) / b (ln )/ + (lnb) /. Hence, the series n diverges. lnn (e) If f(x) then f (x) x(ln x)[ln(ln x)] ln(x) ln(ln x)+ln(ln x)+ x [lnx] [ln(ln x)]. When x > e, f (x) < which implies f(x) is decreasing. Because f(x) is positive, x(ln x)[ln(ln x)] continuous and decreasing on [, ), we may apply the integral test: dx x(ln x)[ln(ln x)] [ ln(ln x) Thus, the series x(ln x)[ln(ln x)] converges. (f) Since n < n + n + for all n, we have ] b n + n + < n. ln(ln ) ln(lnb) ln(ln ). Because /n converges (it s a pseries with p > ), the comparison test implies that /(n + n + ) also converges. (g) Since n n < n and n 4 < n 4 + for n, we have n n n 4 + n n 4 n. Since the series /n converges (it s a pseries with p > ), the comparison test implies that the series n n also converges. n 4 +
13 Review: page of MAT V (h) Since n < n + n, we have n + n n + n < n + n n n n + n n n n + Applying the ratio test to the series n n, we have n+ (n+) n ( ) n. (n + ) n n + n n+ n <. n Therefore, the series n converges. Since / <, (/) n is a convergent n geometric series. It follows that ( n + (/) n ) converges. Because ( n + n n (/) n ) converges, the comparison test implies that n+ n also converges. n+ n (i) Since n n +, it follows that ( ) n n and hence the series ( ) n n diverges. n+ n+ (j) Since n < n +, we have ( ) n n n + n n n. n/ Applying the ratio test to the series n n/, we have n+ (n+)/ n (n + ) n/ n + n (n+)/ n <, / n/ and therefore the series n converges. Finally, the absolute convergence test n/ implies ( ) n n also converges. n + (k) If n is a positive integer then cos(nπ) ( ) n. Clearly < (n + ) < and / n / (n + ). / Therefore, the alternating series test implies that the series cos nπ converges. (n+) / (l) Since cos(n), we have ( ) n+ n cosn n n. n Applying the ratio test to the series n n gives (n+) (n + ) (n+) n n n n +n+ n (n + ) n n+ ( n + n ) n+ <, and therefore the series n converges. Hence ( ) n+ n cos n converges by n n absolute convergence test.
14 MAT V Review: page 4 of (m) Applying the ratio test, we have (n+)! (n+) ((n+)!) (n + )! n (n!) (n+)! (n + )! (n+) ((n + )!) n (n!) and therefore the series (n+)! converges. n (n!) (n) Applying the ratio test, we have ( ) n+ 5 (n ) (n+) 4 7 (n ) (n+) ( ) n+ 5 (n ) 4 7 (n ) n + (n + ) <, n + n + <, and therefore the series ( ) n+ 5 (n ) 4 7 (n ) converges.. (a) Applying the ratio test, we have (n+)xn+ n+ nx n n (n + ) x n x n + n x, which implies that this series converges if x < and diverges if x >. Therefore, the radius of convergence is. (b) Using the ratio test gives ( 4)n+ x n+ n+ ( 4)n x n n+ 4 x n + n + 4 x n + n + 4 x, which implies that that this series converges if x < /4 and diverges if x > /4. Thus, the radius of convergence is /4. (c) Applying the ratio test, we have (x )n+ (n+) 4 +6 (x )n n 4 +6 x (n 4 + 6) (n + ) x n 4 n x 4 which implies that that this series converges if x / < / and diverges if x / > /. Hence, the radius of convergence is /. (d) Using the ratio test yields (n+)!xn+ (n+) n+ n!x n n n ( ) x n n n n x. (n + ) n n +
15 Review: page 5 of MAT V Now, if f(n) ( n n+) n then ln f(n) n ln ( n n+) ( n Hence, n + test, we see that ( ln n n+ n ) ( n+ ) ( ) n+ n n (n+) n n n+. ) n eln f(n) e lnf(n) e. Returning to the ratio (n+)!xn+ (n+) n+ e x, n!x n n n which implies that this series converges if x < e and diverges if x > e. Thus, the radius of convergence is e. (e) Applying the ratio test, we have ( )n+ (x ) n+ (n+) ( )n+ (x ) n n x (n + ) n x ( ) n + x, which implies that this series converges if x < and diverges if x >. Therefore, the radius of convergence is. (f) Using the ratio test, we obtain ln(n+)xn+ n+ x ln(n + ) ln(n)xn ln(n) n x n+ n n x, which implies that this series converges if x < and diverges if x >. Thus, the radius of convergence is. (g) Applying the ratio test gives ( )n+ n+ (x ) n+ (n+)! ( )n+ n (x ) n n! x n + x n +, which implies that this series converges for all x. Therefore, the radius of convergence is. (h) Using the ratio test implies x + 5 n+ x + n 5 x +, 5
16 MAT V Review: page 6 of which implies that this series converges if x + < 5 x <. Similarly, the series diverges is x >. Therefore, the radius of convergence is.. (a) Since e x n x n n! + x + x! + x! +, we have n + e.!!! n! (b) Since sin(x) n ( ) n x n+ (n + )! x! + x5 5! 7! + +, we have ( )n + sin().! 5! 7! (n+)! (c) Since x n + x + x + x +, x n we have n /4 (d) Since cos(x) n ( ) n xn (n)! x! + x4 4! x6 6! +, we have + + cos() cos( ).! 4! 6! 4. The second degree Taylor polynomial for a function f about x is given by f() + f () x + f () x. In particular, a f(), b f () and c f (). (a) Since the yintercept is positive, a > ; since f(x) is decreasing at x, b is negative; because f(x) is concavedown at x, c is also negative. (b) Since the yintercept is negative, a < ; since f(x) is increasing at x, b is positive; because f(x) is concaveup at x, c is also positive. 5. The fourth degree Taylor approximation for e h for h near is (a) Evaluating the it, we have e h h h h P 4 (h) + h + h + h! + h4 4!. + h + h + h + h4 h! 4! h h + h + h. h! 4!
17 Review: page 7 of MAT V Clearly, any Taylor approximation of degree at least two will give the same answer. (b) Evaluating the it, we have e h h h h h + h + h + h + h4 h h! 4! h h + h. h! 4! 6 Clearly, any Taylor approximation of degree at least three will give the same answer. 6. (a) Using the binomial series, we have ( ) ( + y + )(! y + ) ( )( y + )( ) y +!! + y 8 y + 6 y + Setting y x gives x x x x +. (b) The geometric series is + x + x x + x + x 4 +. Integrating both sides of this equation with respect to x gives ln( x) C + x + x + x + 4 x4 +. Since ln(), it follows that C. Finally, setting x y, we have ln( y) y y 8 y 4y 4. (c) Since e y + y + y + 6 y + 4 y4 +, setting y z gives e z z + e z z4 6 z6 + 4 z8 + Multiplying both sides of this equation by z, we obtain z e z z z + z5 6 z7 + 4 z9 +. (d) Recall that sin t t t + t5 t7 + and the binomial series implies! 5! 7! ( ) ( + t + )(! t + ) ( ) ( t + )( ) t +!! + t 8 t + 6 t + Multiplying these two power series gives: + t sin(t) (t )( t + t5 t7 + +! 5! 7! t 8 t + ) 6 t + t + t 7 4 t 48 t4.
18 MAT V Review: page 8 of 7. (a) The approximation sin(θ) θ comes from the linear approximation to sin(θ) at θ. When the graph of sin(θ) is below that of θ the approximation is an overestimate and this occurs on the interval < θ <. Similarly, when the graph of sin(θ) is above θ the approximation is an underestimate and this occurs on the interval < θ <. (b) Recall that the following: Suppose f and all its derivatives are continuous. If P n (x) is the nth Taylor approximation to f(x) about a, then f(x) P n (x) M (n + )! x a n+, where M is the maximum value of f (n+) (x) on the interval between a and x. Now, sin θ θ is the nd degree Taylor approximation to sinθ about. Since the third derivative of sin θ is cos θ which is bounded above by, we have sin θ θ! θ 6, on the interval [, ]. 8. Recall that d arcsin(x) dx x. Using the binomial expansion, we have ( )( ( + y) / ) ( n + ) y n + y n! n ( )( ) ) ( n y n n! n ( ) n( 5 7 (n ) ) y n. n n! Setting y x, we obtain x n n 5 7 (n ) x n. n n! Integrating both sides of this equation with respect to x yields 5 7 (n ) arcsin(x) + C x n+. n n!(n + ) Since arcsin(), we have arcsin(x) n n 5 7 (n ) x n+. n n!(n + )
19 Review: page 9 of MAT V 9. The derivatives of (i) (iii) are: (i) (ii) (iii) dy dx A x + Bx and d y dx A x + B x ( ) A x + bx dy dx A cos(x) B sin(x) and d y A sin(x) B cos(x) dx dy dx d y and x + A dx (x + A) /. Hence, (i) is a solution to (c), (ii) is a solution to (a), and (iii) is a solution to (b).. (a) (i) Euler s method for y (sin x)(sin y) starting at (, ) x approximate y y (slope) x sin() sin() (.)..9 sin(.) sin() (.)..9.8 sin(.) sin(.9) (.)..7.7 sin(.) sin(.7) (.).4.54 (ii) Euler s method for y (sin x)(sin y) starting at (, π) x y y (slope) x π sin() sin(π) (.). π sin(.) cos(π) (.). π sin(.) cos(π) (.). π sin(.) cos(π) (.).4 π (b) From the slope field, we see that solution curve passing through (, ) is increasing and that y π is an equilibrium solution to the differential equation.. (a) Separating variables, we obtain e z dz t dt e z t + C z ln ( t C ). Since z(), it follows that () C and C. Thus, the solution to the initialvalue problem is z(t) ln ( t ) ln ( ) t. (b) Separating variables yields dy + t dt t + y y t + K y t+. t +K Since y(), we have + + K and K. Hence, the solution to the initialvalue problem is y(t) t+. t t +t 4
20 MAT V Review: page of Figure. Slope field for y (sin x)(sin y). (c) Separating variables gives dw w θ sin(θ ) dθ w cos(θ ) + C w cos(θ ) C. Since w(), we have C and C. Therefore, the solution to the initialvalue problem is w(θ). cos(θ )+ cos(θ )+ (d) Separating variables, we have du dx dx u x(x+) x x+ ln x ln x + + K u u ln x+ K. x Since u(), it follows that K ln(). Hence, the solution to the initialvalue problem is u(x) ln x+ x + ln() ln x+. x +. (a) Separating variables, we have dr k dt ln R kt + C R R R e kt, where R ±e C. Thus, the general solution to the differential equation is R(t) R e kt.
21 Review: page of MAT V (b) Separating variables gives dp ( dt ln b + ap t + C P b ap a Ke t b ), where K ±e C. Hence, the general solution to the differential equation is P(t) a (Ket b). (c) Separating variables, we obtain cos x dx sinx +ln t t dt ln sin x ln(t) + (lnt) + K sin(x) Cte (ln t), where C ±e C. Hence, ) the general solution to the differential equation is x(t) arcsin (Cte (ln t). (d) Separating variables gives: dx dt ln ln x ln t + C ln x Kt x e Kt, xln(x) t where K ±e C. Therefore, the general solution to the differential equation is x(t) e Kt.. (a) Since the rate of growth of a tumor is proportional to the size of the tumor, we have ds dt ks, where k is a constant. (b) Separating variables, we have ds k dt ln S kt + C S S S e kt, where S ±e C. Therefore the general solution is S(t) S e kt. (c) Since S() 5, it follows that S 5 and S(t) 5e kt is the solution to the initial value problem. (d) Since S() 8, we have 8 5e k which implies k ln ( 8 5).57. Hence, the solution is S(t) 5 ( 8 t/. 5) 4. (a) The volume V of a spherical snowball is V 4 πr where r is the radius. Solving for r gives r ( V 4π ). On the other hand, the surface area S of a sphere is ). Therefore, the differential S 4πr and substitution yields S 4π ( V 4π equation for the volume is dv dt k4π ( ) V, 4π
22 MAT V Review: page of where k is a constant. For simplicity, we combine the constants and write dv dt AV, where A is a constant. Because V is decreasing, A will be negative. (b) Separating variables, we have V / dv A dt V / At + C V ( ) At+C. Since V () V, we have V ( ) C and C V /. Thus, the general solution ( ) A. is V (t) t + V / (c) The snowball disappears when V (t). Solving for t gives ( ) A t + V / t V / A ; note that t is positive because a is negative.
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