Solution of Math132 Exam1

Transcription

1 Solution of Math3 Exam. ( %) (a) A bacteria has population at time t = an its rate of growth is (t 3 + t + ) bacteria per hour after t hours. What is the population after 4 hours? Suppose p(t) is the bacteria population after t hours., then p (t) = (t 3 +t+) is the rate of growth of the bacteria population, then by the net change thm, (b) Let since p() =, then p(4) = p (t)t = p(4) p() (t 3 + t + )t = + 4 (t 3 + t + )t = +( t4 4 + t +t) 4 = +( ) = +76 = 8 Fin f (x). π x ( f(x) = π sin( t) t) = sin( t) t + ex By the chain rule an the FTC, we know that therefore x ex π x. (%)Evaluate r 4 ln r r u(x) a π sin( t) t. f(t) t = f(u(x))u (x) sin( t) t = sin( e x )(e x ) = e x sin( e x ) By the formula of integration by parts uv = uv vu

2 , so set then therefore, 3. (%)Evaluate = r ln r = r ln r u = ln r r 4 r = v v = r r 4 ln r r = ln r ( r ) r r r (ln r) = ln r r 4 r r = (t 6t 6) t ln r r + C t 6t 6 = (t 8)(t + ) r r (t 8)(t + ) = A t 8 + A t + A (t + ) + A (t 8) = A =, A =, therefore, (A + A )t + (A 8A ) = { A + A = ; A 8A =. (t 6t 6) t = ( t 8 t + ) t = t 8 t t + t set u = t 8, then u = t, it follows t 8 t = u u set v = t +, then v = t, it follows t + = v v, thus u u v = ln u ln v + C ln t 8 ln t + + C v

3 sin 3 x 4. (%)Evaluate cos x x solution: sin 3 x (sin cos x x = x)(sin x) cos x ( cos x)(sin x) x = x cos x set u = cos x, then u = sin x x sin x x = u, therefore, =. (%)Evaluate ( u )( u) u = u = (u u u u ) u u u u u = u ln u + C = cos x ln cos x + C x ( + x ) 3/ set x = tan θ, since x [, ], then take θ < π, in this case, x sin θ = + x then an ( + x ) 3 = ( + tan θ) 3 = (sec θ) 3 = sec 3 θ x = sec θθ x = ( + x ) 3/ sec 3 θ sec θθ = sec θ θ = cos θθ = sin θ+c = x + x +C ( + x ) 3/ x = x + x = 6. ( %) (a) Fin the area of the boune region enclose by the curves y = x an y = 4x. Fin the intersection points of these two curves. x = 4x x 4x = x( x) = x =, x = Fin the area of the enclose region. 3

4 (x 4x )x = x 4 3 x3 = ( ) 4 3 ( )3 = 4 6 = (b) Fin the volume of the soli obtaine by rotating the boune region in (a) about the x-axis. Fin the outer raius r o, inner raius r i an the crosssectional area A(x). therefore, r o = x r i = 4x. ( %) A(x) = π(r o r i ) = π((x) (4x ) ) = π(4x 6x 4 ) Fin the volume v = A(x)x = π(4x 6x 4 )x = 4π 3 x3 6π x = π 6 π = π. Solution of Math3 MAKE-UP Exam (a) A car travels back an forth along a straight roa. The velocity of the car (in mph) at time t (in hours) is given by the function v(t) = t 9t + 8. Calculate the total istance travelle by the car from t = to t =. v(t) = t 9t + 8 = (t 3)(t 6) v(t) > for t (, 3) an v(t) < for t (3, ) The total istance travelle by the care from t = to t = v(t) t = 3 (t 9t+8)t+ 3 = 6 3 (98 36) = 3 (t 9t+8)t = ( t3 3 9 t +8t) 3 ( t3 3 9 t +8t) 3 4

5 (b) Let Fin f (x). f(x) = et t + x ( by the chain rule an FTC, x ( x. (%)Evaluate 3r e r r x et t) = sin( e t ) t. sin( e t ) t) = (x ) sin( e x ) = x sin( e x ) use the integration by parts, uv = uv vu set u = r, then v = e r r v = e r 3r e r r = 3 r (e r ) = 3r e r 3 e r (r ) = 3r e r 6 re r r use integration by parts again, set u = r, then v = e r r v = e r, then 3r e r 6 r(e r ) = 3r e r 6[re r e r r] = 3r e r 6re r + 6e r + C 3. (%)Evaluate () set x + x 3 x x + x 3 = (x + 6)(x ) x + x 3 = A x A x A (x ) + A (x + 6) = (A + A )x + (6A A ) =

6 therefore, { A + A = ; 6A A =. A =, A = (x + x 3) x = ( x x + 6 ) x = x x x + 6 x set u = x, then u = x, it follows x x = u u set v = x + 6, then v = x, it follows x + 6 = v v, thus u u 4. (%)Evaluate tan 3 x x v = ln u ln v + C ln x ln x C v = tan 3 x x = (tan x)(tan x) x (sec x )(tan x) x = = sec x(sec x tan x x) sec x tan x x sin x cos x x tan xx Use u-substitution, set u = sec x, then u = sec x tan x x, we have sec x(sec x tan x x) = uu similarly, set v = cos x, then v = sin x x sin x x = v, so sin x cos x x = v v = v v, therefore, uu + u v = v + ln v + C = sec x + ln cos x + C 6

7 . (%)Evaluate (x + 4) x 3 set x = tan θ, since x [, ], then take θ < π, in this case, x sin θ = x + 4 then an (x + 4) 3 = (4 + 4 tan θ) 3 = 8(sec θ) 3 = 8 sec 3 θ x = sec θθ x = (4 + x ) 3/ 8 sec 3 θ sec θθ = = 4 sin θ + C = x 4 + C 4 + x 4 sec θ θ = 4 cos θθ ( + x ) x = x 3/ 4 + x = 6. ( %) (a) Fin the area of the boune region enclose by the curves y = x 4 an y = x. Fin the intersection points of these two curves. x = x 4 x x 4 = x( x 3 ) = x =, x = Fin the area of the enclose region. (x x 4 )x = x x = = 3 (b) Fin the volume of the soli obtaine by rotating the boune region in (a) about the y-axis. Fin the outer raius r o, inner raius r i an the crosssectional area A(y). therefore, r o = y 4 r i = y A(y) = π(r o r i ) = π((y 4 ) (y) ) = π( y y ) 7

8 Fin the volume v = A(y)y = π( y y )y = π 3 y 3 π 3 y3 = π 3 8