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Unformatted text preview: MATH 31A (Butler) Practice for Final (A) Try to answer the following questions without the use of book, notes or calculator; but you can use the equation sheet posted on the course website. Time yourself and try to finish the questions in less than three hours. 1. Find Z π π/ 2 sin  θ  dθ . When we have an absolute value it is almost always best to break the function up into pieces. In this case for θ ≤ 0 and θ ≥ 0. In particular we note that sin  θ  = sin( θ ) = sin θ if θ ≤ , sin θ if θ ≥ . So using this we now break up our integral and integrate each part. So we have Z π π/ 2 sin  θ  dθ = Z π/ 2 sin  θ  dθ + Z π sin  θ  dθ = Z π/ 2 ( sin θ ) dθ + Z π sin θ dθ = cos θ θ =0 θ = π/ 2 cos θ θ = π θ =0 = ( cos0 cos( π 2 ) ) ( cos π cos0 ) = 3 . 2. (a) Find d dx ( f ( x ) g ( x ) h ( x ) ) in terms of f ( x ), g ( x ), h ( x ), f ( x ), g ( x ) and h ( x ). This is the product rule, but instead of two functions, now we have three functions! One way to think of this though is to group two of the functions together and then we apply the product rule twice. So we have d dx ( f ( x ) g ( x ) h ( x ) ) = d dx ( f ( x ) ( g ( x ) h ( x ) )) = f ( x ) ( g ( x ) h ( x ) ) + f ( x ) d dx ( g ( x ) h ( x ) ) = f ( x ) ( g ( x ) h ( x ) ) + f ( x ) ( g ( x ) h ( x ) + g ( x ) h ( x ) ) = f ( x ) g ( x ) h ( x ) + f ( x ) g ( x ) h ( x ) + f ( x ) g ( x ) h ( x ) . (Note: this same pattern (where we add up all the ways to take the derivative of a single piece) holds in general for the product of any number of functions.) (b) Starfleet intelligence has recently learned of a new threat. A new Borg vessel has been discovered that can change its shape, they are calling it the B1000 (short for Borg1000). The B1000 though still has some limitations, first the only shape it can have is a three dimensional box and second the volume is always fixed, i.e., the box cannot deviate from a fixed volume. You have been on a shuttle tracking the B1000. From an earlier observation you saw that it had dimensions 20 meters by 15 meters by 10 meters. Currently though you can only see two sides. You notice that the length is currently 12 meters and is increasing at a rate of 1 meter per minute; the width is currently 25 meters and is decreasing at a rate of 2 meters per minute. What is the current depth of the B1000 and how fast is it changing? At first glance, this seems to have nothing to do with part (a), but let us continue and perhaps we will see the connection. First, we note that since the B1000 is always a box that we have V = xyz where x , y and z are the length, width and depth respectively. The fact that the B1000 cannot change its volume tells us that dV dt = 0. From our earlier observation of the B1000 we know that V = 20 · 15 · 10 = 3000. This allows us to easily solve for the depth since we have...
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This note was uploaded on 04/23/2010 for the course MATH 26218120 taught by Professor Butler during the Spring '10 term at UCLA.
 Spring '10
 BUTLER
 Math

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