MATH MID-TERM EXAM SOLUTIONS CLAY SHONKWILER ( Find the area of the region enclosed by the curves y and y. Answer: First, we find the points of intersection by setting the two functions equal to eachother:. Squaring both sides yields +, so, multiplying by and moving it all to one side + 5 ( 5(, so and 5. Therefore, the desired area is given by 5 ( 3 ] 5 ( ( 3/ ( (8 ( 3 3 3. ( Which grows faster, log 3 or log? Answer: To answer this question, we consider the it log 3 log ln ln 3 ln ln ln ln 3 ln ln By L Hôpital s Rule, this, in turn, is equal to ln ln 3 ln ln ln 3 ln 3. Since this it is finite and non-zero, log 3 and log grow at the same rate. (3 Find the arc length of the curve y from to π/. Answer: We use the arc length formula L π/ + ( dy
CLAY SHONKWILER Now, so dy ( dy. Thus, L π/ (, + π/ π/ π/ + sin ] π/. Now, I made a mistake in writing this problem; sin (π/ doesn t even make sense and sin ( isn t eactly easy to figure out. Anyway, I gave full credit for getting to this stage. ( Let f( sec ( sin. Evaluate the integral f(. Answer: Let α sin. Then we can draw a reference triangle with a side of length opposite the angle α and a hypotenuse of length. Then the adjacent side is 3. Thus, cos α 3, so sec α sec ( sin 3. Hence, f( sin 3 ] 3 sin sin (5 Find π π. csc +
Answer: MATH MID-TERM EXAM SOLUTIONS 3 csc + + sin. Now, sin, so we can apply L Hôpital s Rule to see that this it is equal to + sin cos. Again, we have, so we apply L Hôpital s Rule again to get + cos sin + cos sin, since cos. ( Solve the initial value problem dy y, y(. ln y Answer: This is a separable differential equation; we multiply both sides by ln y y to separate the variables: ln y dy. y Now, we integrate: +C. On the other hand, to compute ln y y dy, let u ln y. Then du y dy, so ln y y dy udu u + C (ln + C. Thus, if we let C C C and equate both sides, we see that (ln y + C. Then (ln y + C, so ln y + C, meaning that y e +C. Now, using our initial condition y( e ( +C e +C, so, since e, C, meaning that C. Therefore, y e. (7 Solve the initial value problem dy + y, > y(.
CLAY SHONKWILER Answer: This is a linear first order differential equation; we divide by to put it into standard form: dy + y. Here P ( and Q(. Hence, P ( ln. Since >, we can leave off the absolute values, so we have Hence, y v( v( er P ( e ln. v(q( 3 [ 3 + C ] Now, using our initial condition, so C. Therefore, 3 3 + C y( 3( 3 + C 3 + C, y 3 3 +. (8 Find the volume of the solid obtained by rotating the region bounded by y +, and the - and y-aes about the -ais. Answer: Using the disc method, πr. Now, the radius of the disc is given by y +, so ( π π + + π tan ] ( π π π.
MATH MID-TERM EXAM SOLUTIONS 5 (9 Find the volume obtained by rotating the region described in problem 8 about the y-ais. Answer: Here, we use the shell method πrh. Now, r and h y +, so π π + + Now, let u +. Then du, so π ( Let f( tan ( + du u π [ u ] π( π(.. Find df. π/3 Answer: The easiest way to see this is probably just to solve for f (, which we do by switching the variables and solving for y: ( y + tan. Then tan y+, so y tan. Hence, f ( tan, so Hence, df df sec. (sec π/3 ( 8. π/3 Alternatively, we could use the fact that df df. f(a a Now, df + ( +. + (+
CLAY SHONKWILER Now, note that tan ( 3 π/3. Hence, if a+ 3, then f(a π/3. Solving for a, we see that a 3. Therefore, df 3 + (( 3 + + ( 3 8. Thus, df π/3 df 8 3 8. ( Prove that sec θ tan θdθ sec θ + C. Proof. Note that sec θ tan θdθ cos θ sin θ cos θ dθ sin θ cos θ dθ. If we let u cos θ, then du sin θdθ, so sin θ du cos θ dθ u u + C + C sec θ + C. cos θ ( Let S be a solid whose base is the ellipse + y and whose cross-sections perpindicular to the -ais are right isosceles triangles with one of the equal (i.e. non-hypotenuse sides determined by the ellipse (meaning the other equal side is vertical. Find the volume of S. Answer: First, let s re-write the equation for the ellipse; solving for y, we see that y ±. Hence, the length of the non-hypotenuse sides of the cross-sectional triangles is. Therefore, the area of each cross-sectional triangle is given by A( ( (.
MATH MID-TERM EXAM SOLUTIONS 7 The volume of the solid is thus given by A( [ 3 ( 8 ] ] ( + 8 8 8 8 3 3. (3 Find the volume of the solid obtained by rotating the region bounded by y, and y about the line. Answer: Using the shell method, πrh. Now, since the ais of rotation is, r, while h (since we want the height of the region between y and y. Thus, π( ( π [ 8 + 3] π [8 3 3 + [ π 8 ] 3 + [ π ] 3 [ ] π 3 π 3. ( Suppose you invested $ in an account in which interest is continuously compounded. After year, the amount now in your account is $. Assuming you never withdraw any money from it, in how many years will you have accumulated $, in the account? ]
8 CLAY SHONKWILER Answer: We know that continuously compounding interest is modeled by the formula A(t P e rt. Now, since you start the account with $, P. Now, after year, the scenario is that A(t e r( e r. Dividing both sides by, we see that er. Taking the natural log of both sides tells us that r ln ln ln. Now we re ready to answer the question: the account will have $, in it when Dividing by,, A(t e (ln ln t. e (ln ln t. Taking the natural log of both sides, we see that ln (ln ln t, so ln t. years, ln ln which is option (d. (5 Find the surface area of the solid generated by rotating the region enclosed by the curves y, y and 3 about the -ais. Answer: Note that if y, then, so the surface area is given by ( dy SA πy +. Now, so ( dy. Hence, SA dy, π ( + π + π +.
MATH MID-TERM EXAM SOLUTIONS 9 Now, if we let u +, then du, so we can re-write the integral as ] SA π udu π 3 u3/ 8π 5π [8 ] 3 3. DRL 3E3A, University of Pennsylvania E-mail address: shonkwil@math.upenn.edu