MAT 111 Practice Test 2

Transcription

1 MAT 111 Practice Test 2 Solutions Spring

2 1. 10 points) Fin the equation of the tangent line to 2 + 2y = 1+ 2 y 2 at the point 1, 1). The equation is y y 0 = y 0) So all we nee is y/. Differentiating the above equation implicitly, we get So, substituting = 1 an y = 1, we get 1 2 y y ) = 2y2 + 2y y 2 1 y ) = y solving, we get y = 1. So the equation of the tangent is y 1 = 1 1) OR y = + 2 2

3 2. 30 points) Evaluate the following: a) lim Solution: Notice that if we let f) = 4 2 then f2) = 12 an so the above limit is actually equal to f) f2) lim 2 2 which is precisely f 2). Thus, all we nee to o is evaluate f ) an substitute 2. f ) = an so f) f2) lim = f 2) = = For the PE class stuents, the same solution with f) = an f ) = an limit = 34). b) Fin f ) if f) = sec 3 1+sin Solution: By Quotient Rule, f ) = 1 + sin ) sec 3) sec sin ) 1 + sin ) 2 By Chain Rule sec 3)) = sec 3) tan 3) 3) = 3 sec 3) tan 3). So the final answer is: 3 sec 3) tan 3)1 + sin ) cos ) sec 3) i + sin ) 2 c) Fin f ) if f) = sin 2 ) 3 +1 cos +2. Solution: This is just a repeate application of chain rule, quotient rule an prouct rule: Outer most functions is square, net is sin an then there is a quotient to be ifferentiate. f ) = 2 sin ) cos [ sin )] cos + 2

4 = 2 sin ) [ = 2 sin )] cos cos + 2 cos + 2 ) [ cos cos + 2 cos + 2 { } cos + 2 )] { cos + 2) 3 + 2) cos + 2) 2 } cos + 2) ) [ = 2 sin )] cos cos + 2 cos + 2 { cos + 2) )) 3 } + 2 sin ) cos + 2) 2 ) = 2 sin ) cos cos + 2) ) sin 3 +1 cos + 2 cos + 2 cos + 2) 2 ) = 2 sin ) cos cos + 2) ) + sin cos + 2 cos + 2 cos + 2) 2 4

5 3. 25 points) Assuming that the earth is locate at the point 2, 0), a comet is observe to be moving along a parabolic path given by y = when the istance is measure in units of million km). It is observe that the co-orinate of the comet is changing at 3000 km/s. a) How fast is the istance between earth an the comet changing, when the comet is at 1, 1). b) A telescope on earth is constantly tracking the comet. How fast is the telescope rotating, when the comet is at 1,1)? This problem is almost ientical to Problem 20 from Section 2.8 which was on the homework. Solution: Part a): The location of the comet is, y), but since it moves on the curve y =, it s location is, ). So, its istance D from earth s location 2, 0) is D = 2) 2 + 0) 2 = = Differentiating, we get D = = ) Now, if istances are measure in units of million km, then we have to convert the rate of change of into that same unit. So, = 3000/106. So, substituting = 1 an = 3000/106, we get D = ) = million kilometers/sec

6 In other units: D = km/s Part b) Now, we just have to relate the angle with quantities for which we o know rate of change. We have cosθ) = 2 D Now, we have two choices. We can o implicit ifferentiation, an substitute later using D/ calculate in part a), or we can substitute for D in terms of an then just ifferentiate. We will o the first metho, since it s less messy. We get: sinθ) θ D D 1) 2 ) = D 2 When the comet is at 1, 1), we have D = 2 an θ = π/4 so sin θ = 1/ 2 an = 3000/10 6. We also know from part a) that at this point D/ = Substituting, we have 1 θ 2 = ) ) 2 So, simplifying, we get. θ = raians/s 6

7 4. 25 points) A runner Let s call him C) is running aroun a 400m running track a real one this time), shown in the figure below, at 7m/s. His coach Tom is staning 11 meters away from the nearest point on the track as shown in the figure, recoring the run on a hanicam for analysis later an for posting on youtube). Assume that 100/π) 33 for ease of calculation) an answer the following. a) How fast is the istance between Tom an C changing, when C is 33 meters away from Tom. This is the rate at which the zoom on the camera nees to be ajuste to maintain a full frame close up of C). b) How fast shoul Tom be turning to keep C in frame, when C is at the point P shown. This problem is really two problems that were iscusse in class. Part a) is the same as Problem 37 in Section 2.7 an Part b) is the same as Problem 26 from the same section. Solution: First we nee to figure out some geometric information. Accoring to the problem the total length of the track is 400m an the straight parts of the track are 100m each. So the remaining two semi-circular parts are also 100m each. So we have πr = 100 an so r = 100/π 33. Part a): Since r = 33m, the istance between Tom T for short) an the point Q which in the figure the center of the semi-circle) is 44m. So, using the law of cosines for triangle TCQ, we can write So ifferentiating, we get D 2 = cos θ 2D D = sin θ)θ = sin θ)θ 7

8 Now, since C is running at 7m/s, we can calculate θ/. It takes him 100/7 secons π to run the semi-circle, which is π raians. So the angular spee is 100/7 = 7/33 again using the approimation 100/π 33). Now, all we nee to o is sin θ when D = 33, an we can just substitute. One brut force way is to calculate cos θ using the law of cosines again to fin cos θ an then fin sin θ. But when D = 33, the triangle CQT isosceles, so our life is much easier. If we rop a perpenicular from C onto line QT, it will bisect it into two equal parts QM an MT both of length 22m. Thus, looking at the figure, we have CM = = An so, sin θ = CM CQ = = 5 3. Now, we can substitute everything in the equation above, an we get: An this gives 2 33 D = D = Part b): This part is much easier. We have the easy iagram shown below. We know that the istance T S is increasing at 7m/s an we want to fin θ. 8

9 We know that the istance T S is increasing at 7m/s an we want to fin θ. But we have tan θ = P S/ST an so. Thus, ifferentiating we get: ST = 33 cot θ ST = 33 csc2 θ) θ Now, from the iagram, when C is at P, we have ST = = 77 an so we have P T = = So, csc θ = 11 58/33 = 58/3. Substituting in the above equation we get 7 = 33 θ = 58 3 ) 2 θ ra/s 9

10 5. 10 points) Fin the approimate value of tan51 ) using linear approimation at 0 = 45. In orer to o calculus, we nee convert everything into raians. 45 = π/4 raians an 51 = 51π/180 raians. Now the linear approimation at 0 is L) = f 0 ) + f 0 ) 0 ). In our case, 0 = π/4, f) = tan ) an so f ) = sec 2 ). So, f 0 ) = tan π/4 = 1 an f 0 ) = sec 2 π/4 = 2) 2 = 2 an we have. Substituting = 51π/180 we get L) = π/4) tan 51 ) π 180 π 4 ) = π π 180 ) = π 180 = π 30 So, if we approimate π 3, we get tan51 ) 1 + 2/10 =