1. The following problems are not related: (a) (15 pts, 5 pts ea.) Find the following limits or show that they do not exist: arcsin(x)

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1 APPM 5 Final Eam (5 pts) Fall. The following problems are not related: (a) (5 pts, 5 pts ea.) Find the following limits or show that they do not eist: (i) lim e (ii) lim arcsin() (b) (5 pts) Find and classify all relative etrema of f() = (iii) lim ) +(sin (c) ( pts) A right triangle is formed in the first quadrant by both the -ais and y-ais and a line through the point (, ). Find the vertices of the triangle so that its area is a minimized. [Hint: the slope of the hypotenuse should be the same whether slope is found using the y-intercept and the point (,) or the -intercept and the point (,).] (a)(i) Note that, lim e e (a)(ii) Applying L Hospitals Rule yields, arcsin() L lim H L H e =. (a)(iii) If we let y +(sin ), then, using continuity, = ln y + ln(sin() ) + ln(sin()) = ln(sin ) L lim H cot + / + + tan() Finally, applying L Hospitals Rule again yields so ln(y) = implies y = e =, that is L ln(y) H + tan() + sec () = lim +(sin ) =. Note that we could also do this limit directly using continuity by rewriting the limit as lim +(sin ) + eln(sin()) + e ln(sin()). (b) Taking the derivative yields f () = 5 5 =, so the critical points are = ±. Now we need 5 to classify the critical points. Using, for eample the First Derivative Test, we have st Derivative Test for f(): - Therefore, we have a relative maimum at (, 5 ), and relative minimum at (, 5 ). (Note could also use the nd Derivative Test.) (c) Assume the -intercept and y-intercept of the line are (, y) and (, ) respectively.

2 Note that we must have >, since if = then the line through (, ) and (, ) is vertical and we don t get a triangle, and if < then the line through (, ) and (, ) has a negative y-intercept and we don t get a triangle solely in the first quadrant. Then the slope of the hypotenuse using the two intercepts is y =. This produces the relationship y = +. So we wish to minimize the area A(, y) = y, subject to y = + A() = ( + ) = + Then differentiating and finding critical points yields, A () = + ( ) ( ) = ( ) = and >. Substituting, yields Thus ( ) = implies = ± and so the critical points are =,. But = is not in the domain > so we may ignore it. Note that A () = ( ), and by the nd Derivative test A () >, so we have a relative minimum at = (we may also use the First Derivative Test to determine this). Since = is the only critical point however, it gives an absolute minimum. Thus, select = then y = and A = and the vertices that yield the minimum area are (, ), (, ), (, ).. The following problems are not related: (a) (5 pts, 5 pts ea.) Find dy/ if: (i) y = e +tan () (ii) y = cosh( ) (iii) e y tanh(y) =, 69 (b) (7 pts) Consider the function f() = + on the interval [, ]. (i) ( pts) Does the function f() satisfy the hypotheses of the Mean Value Theorem on the given interval? (ii) ( pts) If it satisfies the hypotheses, find all numbers c that satisfy the conclusion of the Mean Value Theorem. If it does not satisfy the hypotheses, write DNE. (c) (8 pts) A kite feet above the ground moves horizontally at a speed of 8 ft/s. At what rate is the angle between the string and the horizontal decreasing when ft of string has been let out? (a)(i) Using the Chain Rule, we have dy = e+tan () (a)(ii) Using the Product Rule and Chain Rule, we have ( + ) + = e +tan () ( ) + + d ( cosh( ) ) = cosh( ) + sinh( ) = cosh( ) + / sinh( ). (a)(iii) Using implicit differentiation yields e y + e y y sech (y)y = = y = e y sech (y) = e y e y sech (y) e y

3 (b)(i) Yes. The function is a polynomial, so it is continuous on [, ] and differentiable on (, ). (b)(ii) We see that c =, since f (c) = f(b) f(a) b a c = = c = c = which is in (, ). (c) Let θ denote the angle of the kite string with the ground, let denote the horizontal distance of the kite from the person holding the string and let y be the amount of string let out. Then Now, dt cot(θ) = = cot(θ) dt = csc (θ) dθ dt dθ = 8ft/s and when y = we have sin(θ) = =, so so the angle is decreasing at a rate of dθ dt = (θ) sin dt = (/) 8 = 5 rad/s 5 rad/s. dt = (θ) sin dt. The following problems are not related: (a) (5 pts, 5 pts ea.) Evaluate the integrals: (i) e / (ii) + (iii) e e (b) (5 pts) Find h in terms of f and g if h() = f()g() f() + g(). (c) ( pts) The Ideal Gas Law relating the pressure, volume, and temperature of a gas is given by P V = nrt where R is the ideal gas constant. Assuming that the current pressure of the gas, P, is atmospheric pressure units (atm) and assuming that n, R, and T are held fied, use differentials to estimate the relative change in the volume, V, if the pressure increases by. atms. (a)(i) Let u = then du = and e / = /8 e u du = /8 e u du = [ e u ] /8 = (e e/8 ) (a)(ii) Factoring out a from the denominator yields + = + = (/) + let u = /, then du =, and so we have (/) + = (a)(iii) Let u = e, then du = e and e = e du u + = tan (u) + C = ( ) tan + C du u = sin (u) + C = sin (e ) + C

4 (b) Use the Quotient Rule and Product Rule, h = (f g + fg )(f + g) fg(f + g ) (f + g) = ff g + f g + f g + fgg ff g fgg (f + g) = f g + f g (f + g) (c) We have Then P V = nrt V = nrt P dv V = nrt dp P nrt P dv = nrt P = P dp = (.) =. dp. The following problems are not related: (a) (5 pts, 5 pts ea.) Find dy/ given: (i) y = (sin ) (ii) y = + (iii) y = e u() where u() = g() cos(f(t)) dt (b) (5 pts) In your blue book clearly sketch the graph of a function f() that satisfies the following properties (label any etrema, inflection points or asymptotes): f() =, f( ) = f() lim f() =, lim f() = + lim f() = f () > if < f () < if < (c) ( pts) A bacteria culture initially contains 6 cells and its population grows at rate proportional to its size. After an hour the population has increased to. Find an epression for the number of bacteria after t hours. (a)(i) Using logarithmic differentiation yields, ln(y) = ln[sin() ] = ln(y) = ln[sin()] = y y = ln[sin()] + cos() sin() and so y = (sin()) [ ln(sin()) + cot() ]. (a)(ii) Applying the Quotient Rule yields, dy = ln() ( + ) ln() ( + ) = ln() ( + ) (a)(iii) First note that, dy/ = e u() u () by the Chain Rule, and applying the Fundamental Theorem of Calculus to determine u (), we get dy = eu() u () = e u() (cos(f(g())) g () ) = e u() g () cos(f(g())) (b) There are vertical asymptotes at = ±, a horizontal asymptote at y = and a local maimum at (, ), the graph could, for eample, look like:

5 (c) Let y(t) represent the population of the bacteria in cells at time t hours, then dy = ky for some dt constant k, and so y(t) = y e kt where y = y() = 6. Now note that y() = implies = y() = 6e k = e k = ( ) = k = ln 6 6 ln and so y(t) = 6e ( ) t 6 = 6 ( ) t The following problems are not related: (a) (5 pts, 5 pts ea.) Evaluate the integrals: ( 6 (i) + sin() ) + (ii) (b) (5 pts) Use the Squeeze Theorem to show lim + sin + (iii) + ( ) =. ln( e ) (c) ( pts) The equation of motion of a particle is s(t) = t t 6t ft (i ) Find the velocity at time t. (ii ) When is the particle moving in the positive direction? (iii ) Find the acceleration after seconds. (iv ) Find the total distance traveled during the first seconds. (a)(i) First note that y = sin() is an odd function, and using geometry to calculate the first integral we + get, ( 6 + sin() ) + = 6 sin() + + = 6 + = π = 8π (a)(ii) Breaking up the integral yields, + + = + + +

6 Now, in the second integral, let u = + then du =, and + + (a)(iii) First note that + = tan ()+ ln( e ) = e ln( e ) = e Alternately, one could also use the substitution u = ln( e ). u du = tan ()+ ln u +C = tan () + ln( + ) + C ln(), now let u = ln() then du = and so ln() = du e u = e ln u + C = ln ln() + C e (b) Note that sin ( ( ) implies sin ) and lim ± = implies that ( ) + lim sin = by the Squeeze Theorem. + (c)(i) Here v(t) = s (t) = t t 6. (c)(ii) Note, v(t) = t t 6 = (t )(t + ) and v(t) > when t >, so the particle is moving in a positive direction after eactly seconds. (c)(iii) Note that a(t) = v (t) = t and so a() = 6 = 5 seconds. (c)(iv) The total distance in the first seconds is v(t) dt = v(t) dt + ] v(t) dt = [ [ ] t = t t 6t + t 6t [ ] [ 6() 6() (t t 6)dt + (t t 6)dt [ ] = 6() ] = + = 9 ft. [ ] 6()