Time limit is 80 min.

Transcription

1 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page of 6 Your Name / Adınız - Soadınız Your Signature / İmza Student ID # / Öğrenci No Professor s Name / Öğretim Üesi Your Department / Bölüm Calculators, cell phones off and awa!. In order to receive credit, ou must show all of our work. If ou do not indicate the wa in which ou solved a problem, ou ma get little or no credit for it, even if our answer is correct. Show our work in evaluating an limits, derivatives. Place a bo around our answer to each question. Use a BLUE ball-point pen to fill the cover sheet. Please make sure that our eam is complete. Time limit is 8 min. Do not write in the table to the right. Problem Points Score 8 5 Total:. a) Points Find the equation of the line tangent to the curve / + / = 7 at the point P,). Solution: Assuming that this equation defines implicitl as a function of, first differentiate the given equation with respect to to find the slope of this tangent. d d d /) + d d d / + /) = d d 7) /) = / / d + d = d d = / / = slope = m = / / Hence the equation for the tangent line is = ) 6 = + = 7. [ ] [ ] d = / d,)=,) / = )/ ) / =,)=,) P,) / + / = 7 p., pr. b) Points For what values of c is the curve = c tangent to the line through the points,) and 5,)? +

2 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page of 6 Solution: Let L be the line through,) and 5,). If m is the slope for L, then m = =. Hence L has equation 5 = ) ) = +. Net, we compute d ) c = c d + + ). Suppose L is tangent to curve at =. Then this implies [ c ] c + ) = m = + ) = + ) = c + = ± c = ± c c Net the point A, + ) is on the curve. Similarl the point B, + ) lies on L. Ever tangent must meet the curve at =. Therefore A = B. Hence as c, as otherwise the curve is = and cannot be tangent L, we have c + = + + ) + ) = c Suppose now = + c. Then + = c and + = c. Hence + ) + ) = c c c) = c c c) = c c c = c c = c = c = Similarl, suppose = c. Then + = c and + = + c. Therefore + ) + ) = c c + c) = c c + c) = c + c = c c = which is impossible. Hence the onl possible value is c =. = + 5 p.87, pr.. a) Points Find the volume of solid generated b revolving the region bounded b =, and = about the line =. If, then a vertical strip of the given region at has length ) and moves around a circle of radius, so the volume generated rotation of that region around = is b ) ) shell shell V = π d = π ) )) d = π ) ) d a radius height Solution: = π + )d = π + )d [ = π ] + = π + 8) = π 6 ) = 8π = = p.5, pr.c) b) Points Find the length of the curve = / from = to = 8.

3 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page of 6 Solution: First note that = / d d = ) d / = /. d 9 Therefore 8 ) d 8 L = + d = d + 8 d = 9/ = 9 / + / d 8 9 / + /) d. Now let u = 9 / + and so du = 6 / d. When =, we have u = and when = 8, we have u =. Hence the curve has length L = 8 u / du = [ ] 8 u/ = [ / /] p.77, pr.. a) Points Find the value of lnlog d. Solution: Let u = log and so du = ln) d. When =, we have u = and when =, we have u = log = log =. Hence lnlog d = ln) log ln) d = ln) udu [ ] = ln) u = ln) p.7, pr.b) b) 8 Points Find the limit lim /ln. Solution: The limit leads to the indeterminate form. Let f ) = /ln. Then ln f ) = ln =. Therefore ln lim /ln = lim f ) = lim e ln f ) = e = e p., pr.65a) 9 5 ) /. a) 8 Points Evaluate the integral d.

4 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page of 6 Solution: Let u = 5 and so du = d. When =, we have u = and when = 9, we have u =. Hence 9 5 ) / 9 5 ) / d = = u / du = + [ u / / ] d = / / ) = ) p., pr.65a) b) Points Find the total area of the shaded region. Solution: Let R denote this region. R = {,) R, } {,) R, } }{{}}{{} R Notice that R = R R and R R = /. Therefore AR) = AR ) + AR ). This shows that we will need two integrals. AR) = AR ) + AR ) = ) = + d + [ ] [ = + + ] + = 6 ) ) + = + = 8 ) ) d + ) + d + ) R ) ) d ) 6 8 p.98, pr.6

5 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page 5 of 6 5. a) Points Figure shows two right circular cones, one upside down inside the other. The two bases are parallel, and the verte of the smaller cone lies at the center of the larger cone s base. What values of r and h will give the smaller cone the largest possible volume? p.58, pr.56 Solution: Let h be the height and r be the base radius for the inner cone. From the figure we get r h = 6 r + h =. Now we need to find for what values of h and r, the inner cone will have largest volume i.e., we need to maimize V = πr h. We need to find the absolute maimum value of V r) = πr r) = π r 6 r) over [,6]. Take the derivative and set equal to zero to find the critical points). V r) = π ) r6 r) + r ) = π π r r ) = r r) = r =,r = critical points r r r ) = We have: V ) = ABS. MIN. V ) = π )6 ) = 6π V 6) = p.58, pr. ABS. MIN. ABS. MAX.

6 Januar 5, 7 [: pm-:5 pm] Math / Retake Eam --α-) Page 6 of 6 b) Points Using the following properties of a twice-differentiable function = f ), sketch its graph on the grid b indicating all significant points. Write the intervals of increase/decrease and concave up/down. Solution: The given table indicates that the graph is increasing and concave down on,) has a local maimum at,) decreasing and concave down on, ) has a point of inflection at,) decreasing and concave up on,) has a local minimum at,) is increasing and concave up on,) has a point of inflection at,) Derivatives / Türevler = >, < < >, < =, < < < <, > <, = < < <, > =, > < < >, > >, = < < >, < =, < > <, < = <, < is increasing and concave down on,) has a local maimum at,) decreasing and concave down on,+ ) p.5, pr.