Math 180, Exam 2, Spring 2013 Problem 1 Solution

Transcription

1 Math 80, Eam, Spring 0 Problem Solution. Find the derivative of each function below. You do not need to simplify your answers. (a) tan ( + cos ) (b) / (logarithmic differentiation may be useful) (c) + y = y (here y is an implicit function of ). Solution: (a) We must use the Chain Rule and the fact that d d tan () = + This gives us f () = f () = ( + cos()) + d ( + cos()) d ( sin()) ( + cos()) + (b) We begin by letting y = /. Then taking the natural logarithm of both sides of this equation and using the fact that ln(a n ) = n ln(a) we obtain ln(y) = ln ( /) ln(y) = ln() ln(y) = ln() Net we use implicit differentiation to obtain an equation involving dy. In order to do d this, we must use the Chain and Quotient Rules. d d ln(y) = d ( ) ln() d y dy d = d d ln() ln() d d y dy d = ln() y dy d = ln()

2 Finally, we use algebra to find dy d and then write our answer in terms of only. dy d = y ln() dy d = / ln() (c) Here we use implicit differentiation. Using the Power and Chain Rules we obtain Now we use algebra to find dy d. d d + d d y = d d y + y dy d = dy d y dy d dy d = dy ( y ) = d dy d = dy d = y y

3 Math 80, Eam, Spring 0 Problem Solution. For the function given by f() = answer the following questions: (a) Find the interval(s) where f() is increasing, decreasing. (b) Identify all local etrema. (c) Find the interval(s) where f() is concave up, concave down. (d) Identify all inflection points. (e) Sketch a graph of f consistent with the information determined in parts (a) and (b). Your graph does not have to be precise. Solution: (a) To find the intervals of monotonicity we begin by finding the critical points of f. Since f is a polynomial these points will occur whenever f () = 0. f () = = 0 4 ( ) = 0 = 0, = We now split the domain of f into the intervals (, 0), (0, ), (, ) and let c =,, be test points in each interval, respectively. We then evaluate f (c) to determine if f is increasing or decreasing on each interval. Our results are summarized below. Interval Test Number, c f (c) Sign of f (c) Conclusion (, 0) 8 decreasing (0, ) decreasing (, ) 6 + increasing (b) From the table above we see that, although f (0) = 0, there is no sign change in f across = 0. Thus, f(0) is neither a local minimum nor a local maimum. On the other hand, f () = 0 and f changes sign from to + across = which means that f() = is a local minimum of f according to the First Derivative Test.

4 (c) To find the intervals of concavity we begin by finding the possible inflection points of f. Since f is a polynomial these points will occur whenever f () = 0. f () = 0 8 = 0 4( ) = 0 = 0, = We now split the domain of f into the intervals (, 0), (0, ), (, ) and let c =,, be test points in each interval, respectively. We then evaluate f (c) to determine if f is concave up or concave down on each interval. Our results are summarized below. Interval Test Number, c f (c) Sign of f (c) Conclusion (, 0) 0 + concave up (0, ) concave down (, ) 4 + concave up (d) Since () f (0) = 0 and f ( ) = 0 and () f changes sign across both = 0 and =, we say that = 0, are inflection points.

5 y 0 (e)

6 Math 80, Eam, Spring 0 Problem Solution. Use a linear approimation to estimate the following quantities. In each case indicate whether your answer is an underestimate or an overestimate. (a) ln(0.98) (b) sin(0.0) Solution: (a) Let f() = ln() and a = (this is the number closest to 0.98 for which f(a) is an integer.) The linearization of f at = a is obtained via the formula L() = f(a) + f (a)( a) We know that f() = ln() = 0, by definition. Furthermore, since f () = that f () =. Thus, the function L() is we know L() = 0 + ( ) = The approimate value of ln(0.98) is then ln(0.98) L(0.98) ln(0.98) 0.98 ln(0.98) 0.0 The function f() is concave down for all > 0. This is apparent because f () = < 0 for all > 0. Therefore, we know that the graph of the tangent line, y =, will lie above the graph of y = ln() at = Thus, our estimate is an overestimate. (b) Let f() = sin() and a = 0 (this is the number closest to 0.0 for which f(a) is an integer.) We know that f(0) = sin(0) = 0, by definition. Furthermore, since f () = cos() we know that f (0) = cos(0) =. Thus, the function L() is L() = 0 + ( 0) = The approimate value of sin(0.0) is then sin(0.0) L(0.0) sin(0.0) 0.0 The function f() is concave down on the interval (0, π). This is apparent because f () = sin() < 0 for all in (0, π). Therefore, we know that the graph of the tangent line, y =, will lie above the graph of y = sin() at = 0.0. Thus, our estimate is an overestimate.

7 Math 80, Eam, Spring 0 Problem 4 Solution 4. A rectangle has dimensions cm by cm. The sides begin increasing in length at a constant rate of cm/s. At what rate is the area of the rectangle increasing after 0s? Solution: Let and y be the sides of the rectangle. The area of the rectangle is then A = y Differentiating each side with respect to t we obtain d A = d ( ) y da = d y + dy The rate of change of the length of each side of the triangle is cm/s. Thus, da = (y + ) In order to find the value of da after 0s, we need to know the lengths of the sides of the rectangle after 0s. Since the sides increase at a constant rate, we know that the size of the rectangle is ( + 0) cm by ( + 0) cm after 0 s. That is, the size is cm by cm. Therefore, the rate of change of the area of the rectangle is da = ( + ) = 90

8 Math 80, Eam, Spring 0 Problem 5 Solution 5. Find the point on the graph of y =, > 0 that is closest to the origin. Hint: Use the square of the distance between (0, 0) and (, ) as the function to be minimized. Solution: Using the hint, we define the function to be minimized as f() = + y The constraint in the problem is the equation for the curve. That is The corresponding y-coordinate is y = =. This point corresponds to an absolute minimum value of f because it is the only critical point and as one moves away from this point along the curve, the distance increases. y = Plugging this equation into the equation above gives the function f() = + 4 whose domain is > 0. The critical points of f are points where f () = 0. f () = 0 8 = 0 y (, y) 4 = 4 = (0, 0)