MA Midterm Exam 1 Spring 2012

Transcription

1 MA Miterm Eam Spring Hoffman. (7 points) Differentiate g() = sin( ) ln(). Solution: We use the quotient rule: g () = ln() (sin( )) sin( ) (ln()) (ln()) = ln()(cos( ) ( )) sin( )( ()) (ln()) = ln() cos( ) sin( )( ) (ln()) = ln() cos( ) sin( ) (ln()). (7 points) Differentiate f() = e sin(ln()). Solution: We use the prouct rule an several chain rule applications: ( ) f () = (esin(ln()) ) + e sin(ln()) ( ) = e sin(ln()) (sin(ln())) + esin(ln()) = e sin(ln()) cos(ln()) (ln()) + esin(ln()) = e sin(ln()) cos(ln()) + esin(ln()) = e sin(ln()) cos(ln()) + e sin(ln()).

2 MA Miterm Eam Spring (. (9 points) Differentiate f() = ln e cos () iscusse in class an simplify completely. Solution: Using the Rules of Logarithms: ( ) f() = ln e cos () Now, we ifferentiate: ). You must use the Rules of Logarithms = ln( ) ln(e cos ()) = ln( ) ( ln(e ) + ln(cos ()) ) = ln() ln(e) ln(cos()) = ln() ln(cos()). f () = ( ln() ln(cos())) = ( ln()) () ( ln(cos())) = ln cos() (cos()) = ln + sin() cos() = ln + tan().. (7 points) The number of bacteria N in some particular culture after t minutes is moele by N = e.5t. Fin the rate of change to one ecimal after 5 minutes. Solution: Fining a rate of change involves fining the erivative. t (N) = e.5t (.5) = e.5t. This formula gives the rate of change at any time t. We are aske about t = 5, so we plug e.5 5 = e.5 into our calculator to get.8.

3 MA Miterm Eam Spring 5. (8 points) Differentiate y = sin. Solution: First, we take the natural log of both sies: ln(y) = ln( sin ). Using the Power Rule for Logarithms, ln( sin ) = sin() ln(). Now we hit both sies with, remembering that y is a function of, so we have to use a chain rule: ln(y) = (sin() ln()) (y) = sin() (ln ) + ln (sin ) y y y = sin() + ln cos() () y y = sin() + ln() cos() [ ] sin() y = y + ln() cos() [ ] sin() y = sin + ln() cos(). The last line use the orignal statement of the problem that y = sin. 6. (9 points) Fin the minimum an maimum of the function f() = πe, or write NONE where applicable. State both where an what the etreme values are, that is, the an f() values. Solution: We can only have a min/ma if the erivative is zero, so we ifferentiate using the prouct rule: f () = π( e ) + e (π) = πe ( ). Setting this equal to zero, we notice the first factor is never zero (because eponential functions are always positive). The secon factor is zero precisely for =. Thus, our only caniate for a min/ma is =. To see which it might be, we check the sign (positive or negative) of the first erivative to the left an right. To that en, f () = πe ( ) = π > an f () = π <. Since the first erivative goes from positive to negative, the e function goes from increasing to ecreasing, so = is a maimum. We plug it in to fin the actual value f() = πe. Therefore, the final answers are Maimum: ( ), π e an Minimum: NONE.

4 MA Miterm Eam Spring 7. (8 points) Fin the area between the the y-ais, the -ais, the curve y = ( + ) an the line =. Epress your answer as a fraction. Solution: Area uner a curve (assuming the curve lies above the -ais) is given by an integral. The escrption of the problem means we have to solve u = +. Then u =, so we have Note the enpoint shift using u = +. ( + ) = u u = u = = 6 = 5. ( + ). Let 8. (9 points) Suppose the current in an AC circuit at time t is given by i(t) = cos(t) + sin(t). Fin the first maimum after t =. Only an eact answer will receive full creit. You o not have to specify which t value correspons to the maimum. Solution: Since we are fining a maimum, we ifferentiate: i (t) = sin t + cos t. Setting this equal to zero yiels sin t = cos t, which, after a few ivisions, gives tan t =. Since we are only consiering the first maimum, we assume that there is a unique t value satisfying tan t = an that this t value gives us the maimum current. We will enote it t ma. If we think of t ma as an acute angle on a right triangle, tan t ma = means that the opposite sie ivie by the ajacent sie must be. Using the Pythagorean Theorem, the hypotenuse of this triangle is. Using the same triangle an SOHCAHTOA, we have sin t ma = an cos t ma =. Now we can plug in to the orignal equation: i ma = cos t ma + sin t ma = + = + 9 = =.

5 MA Miterm Eam Spring 9. (8 points) Evaluate the integral: ln Solution: Let u = ln, so u =. Then. (9 points) Evaluate the integral: ln uu = = u / u = u/ / + C + C (ln )/ = + C. = u/ e r tan (e r ) sin r. (e r ) Solution: Let u = e r, so u = e r u = er r. Then e r tan (e r ) sin r = tan (u) (e r ) sin (u) u = sin u cos u sin u u = sin u cos u sin u u = cos u u = sec uu = tan u + C = tan(er ) + C. 5

6 MA Miterm Eam Spring. ( points) Integrate: e. Epress your answer as an integer. + e Solutions: Let u = + e. Then u = u =. Using this, e e e+e + e = = +e e e u u u u e = ln u e = (ln e ln e) = ( ) =.. (9 points) Integrate: Solution: π/8 e sin() cos(). Do not roun your answer. Let u = sin(), so u = cos() u = cos(). Then π/8 e sin() cos() = = sin( π/8) sin( ) = eu e u u = (e e ) = e. Remember that e = ; in fact, for any a >, a, a =. e u u 6