Math 134 Exam 2 November 5, 2009

Transcription

1 Math 134 Exam 2 November 5, 2009 Name: Score: / 80 = % 1. (24 Points) (a) (8 Points) Find the slope of the tangent line to the curve y = 9 x2 5 x 2 at the point when x = 2. To compute this derivative we use the quotient rule: dy = (5 x2 ) d If we evaluate dy/ at x = 2 we obtain: [ 9 x 2 ] (9 x 2 ) d [ 5 x 2 ] (5 x 2 ) 2 = (5 x2 )( 2x) (9 x 2 )( 2x) (5 x 2 ) 2 dy (5 4)( 4) (9 4)( 4) = 1 2 = = 16 1 (b) (8 Points) Compute the derivative of the following function: f(x) = sin ( πx 4 tan(2x) ) To compute this we must use the chain and the product rules: f (x) = cos ( πx 4 tan(2x) ) d [ πx 4 tan(2x) ] [ πx 4 ] tan(2x) + πx 4 d ) [tan(2x)] = cos ( πx 4 tan(2x) ) ( d = cos ( πx 4 tan(2x) ) ( 4πx 3 tan(2x) + 2πx 4 sec 2 (2x) )

2 2 (c) (8 Points) Compute the derivative of the following function: ( ) g(t) = cos t 3 e 1 t +sin(t) Before beginning I will rewrite the function as: ( ) g(t) = cos t 3 2 e t 1 +sin(t) To compute this derivative we start off using the product rule and then we have to use the chain rule several times giving: g (x) = d ] [cos(t 3 2 ) = sin ( t 3 2 e t 1 +sin(t) + cos(t 3 d 2 ) [e ] t 1 +sin(t) ) ( ) e t 1 +sin(t) + cos t 3 2 e t 1 +sin(t) ( t 2 + cos(t) ) ) ( 3 2 t 1 2 = 3 t 2 sin( t 3 )e 1 t +sin(t) + ( 1t 2 + cos(t) ) cos( t 3 )e 1 t +sin(t)

3 3 2. A certain drug is administered to a patient at 12:00 noon via an IV drip. The amount of the drug in the body slowly increases until the drip is empty at which point the amount of the drug in the body decreases as it is processed by the body. Let A be the amount of drug in body measured in ounces and let t be time measured in hours since the drug was administered. Suppose that A is given as a function of t by: A(t) = 4t 2 e ( t 2) (a) (2 Points) Suppose that it is known that A (6) = What are the units of this number? The units of this derivative are ounces per hour or ounces hour. (b) (4 Points) Describe in terms understandable to someone who knows no Calculus what A (6) = means. This quantity represents the instantaneous rate of change of the amount of drug in the body with respect to the time in hours exactly 6 hours after the drug has been administered. Practically this means that if the time changes from 6 to 7 hours since the drug was administered then the amount of drug in the body will decrease by approximately ounces. (c) (8 Points) At what time is the maximum amount of the drug in the body and what is the amount at this time? To find this maximum we first take the derivative of the function with respect to time. This gives: A (t) = 8te ( t 2) 2t 2 e ( t 2) = ( 8t 2t 2 ) e ( t 2) Now to find when the maximum occurs we set the derivative equal to 0 giving: 0 = ( 8t 2t 2) e ( t 2) or 2t(4 t) = 0 Thus at t = 4 hours past noon, the maximum amount of the drug is in the body. The actual amount can be computed by calculating: A(4) = 4(16)e ounces

4 4 3. (8 Points) Use the limit definition of the derivative to verify the formula: d [ ] 1 2 x = 2 2 x The limit definition of the derivative tells us that: substituting f(x) = 2 x we have: f (x) h 0 f(x + h) f(x) h d [ ] 2 x h 0 ( h 0 2 (x + h) 2 x h 2 (x + h) 2 x) ( 2 (x + h) + 2 x) h ( 2 (x + h) + 2 x) 2 (x + h) (2 x) h 0 h( 2 (x + h) + 2 x) 2 x h 2 + x h 0 h( 2 (x + h) + 2 x) h h 0 h( 2 x + h + 2 x) = 1 1 = 2 x x 2 2 x

5 5 4. (10 Points) The figure below contains the plot of the curve described by (x + y 2 ) 2 = 9xy The tangent line to the curve at the point (2, 2) is also plotted. Find the equation of that tangent line. 4 3 y (2,2) x To begin this problem we use implicit differentiation to differentiate the give equation. This gives: Distributing we obtain: Subtracting and adding lead to: 2(x + y 2 )(1 + 2yy ) = 9y + 9xy 2(x + y 2 ) + 4y(x + y 2 )y = 9y + 9xy 4y(x + y 2 )y 9xy = 9y 2(x + y 2 ) Factoring out a y on the right hand side gives: Dividing on both sides gives: y ( 4y(x + y 2 ) 9x ) = 9y 2(x + y 2 ) y = 9y 2(x + y2 ) 4y(x + y 2 ) 9x Now we evaluate at x = 2 and y = 2 which gives: y = 18 2(2 + 4) = 8(2 + 4) = 6 30 = 1 5 The equation of the line can then be found by various ways using y = m = 1 5 point (2, 2). For instance the point slope formula gives: and the y 2 = 1 (x 2) 5 or y = 1 5 x or y = 1 5 x + 8 5

6 6 5. (10 Points) At noon ship A is 150 km east of ship B. Ship A is sailing north at 35 km/h and ship B is sailing south at 25 km/h. How fast is the distance between the ships changing at 4PM? A z 150 km x z x+y y B 150 km When looking at this problem we first define two quantities: Now we know the following rates: x = The distance that ship A has traveled y = The distance that ship B has traveled dt = 35km hr Now let, dy dt = 25km hr z = The distance between the two ships If we draw a picture (see the red triangle above) and use the Pathagorean Theorem then we can recognize that we have the relationship: (x + y) = z 2 If we differentiate with respect to t we get: ( 2(x + y) dt + dy ) = 2z dz dt dt If I solve for dz dt we get: dz ( ) (x + y) dt = dt + dy dt z Now at 4 PM, 4 hours have passed x = 4 35 = 140 km and y = 4 25 = 100 km.. We can calculate z from: (x + y) = z 2 or = z 2 or z = km. Substituting these values in we obtain: dz dt = ( ) ( ) km h

7 7 6. The function f(x) is graphed in the figure below. The function g(x) and its derivative is described by the table to the right of the graph. Use this information to answer the questions below. 4 x g(x) g (x) f(x) (a) (5 Points) If l(x) = x 2 f(x)g(x), what is l (5)? To solve this problem we first apply the product rule to obtain: Now we evaluate at x = 5 giving: l (x) = 2xf(x)g(x) + x 2 f (x)g(x) + x 2 f(x)g (x) l (5) = 10f(5)g(5) + 25f (5)g(5) + 25f(5)g(5) We then have the read the values of f(5) = 1, f (5) = 2), g(5) = 4.2 and g (5) =.4. Thus: l (5) = 10(1)(4.2) + 25( 2)(4.2) + 25(1)(.4) = 158 (b) (5 Points) If h(x) = f(g(x)), what is h (6)? To calculate this we must apply the chain rule giving: Substituting 6 we get: h (x) = f (g(x))g (x) h (6) = f (g(6))g (6) From the chart we have that g(6) = 4.5 and that g (6) =.2. Substituting in these values we have: h (6) = f (4.5)(.2) Now since f has a slope of 2 when x = 4.5 we know that f (4.5) = 2, thus: h (6) = ( 2)(.2) =.4 (c) (2 Points) At what point(s) is it correct to say that f(x) is continuous but not differentiable? The function f is continuous but not differentiable at x = 4 and x = 8. If you included 0 and 10 I did not take off points but technically they are not continuous there, they are only right and left continuous respectively at those points. (d) (2 Points) On what interval(s) is it correct to say that f(x) is increasing at an increasing rate? The function is increasing at an increasing rate in places where it is increasing (going up) and as you move left to right the slope gets steeper and steeper. This is happening only on the interval between (2, 4).