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 cross-sectional 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 cross-sectional 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 cross-sectional 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 cross-sectional 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 cross-sectional 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, left-hand sum will give an underestimate and right-hand sum will give an overestimate. (b) Since the function is concave-down, trapezoid rule will give an underestimate and the midpoint rule will give an overestimate. (c) Since the function is concave-up, trapezoid rule will give an overestimate and the midpoint rule will give an underestimate. Furthermore, because the function is decreasing, left-hand sums will also give an overestimate and right-hand sums will give an underestimate. (d) Since the function is concave-up, 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, arc-length + ( ) 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 ( ) arc-length + 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 arc-length + ( 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 arc-length / / / + ( ) 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 n-th 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 p-series 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 p-series 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 y-intercept is positive, a > ; since f(x) is decreasing at x, b is negative; because f(x) is concave-down at x, c is also negative. (b) Since the y-intercept is negative, a < ; since f(x) is increasing at x, b is positive; because f(x) is concave-up 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 n-th 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 initial-value 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 initial-value 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 initial-value 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 initial-value 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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Math 29 Solutions to Assignment 7. Find the gradient vector field of the following functions: a fx, y lnx + 2y; b fx, y, z x cosy/z. Solution. a f x x + 2y, f 2 y x + 2y. Thus, the gradient vector field
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Solutions to old Exam 1 problems Hi students! I am putting this old version of my review for the first midterm review, place and time to be announced. Check for updates on the web site as to which sections
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