Math 111D Calculus 1 Exam 2 Practice Problems Fall 2001

Transcription

1 Math D Calculus Exam Practice Problems Fall This is not a comprehensive set of problems, but I ve added some more problems since Monday in class.. Find the derivatives of the following functions a) y = x 3 + 6x Answer: y = 3x + 6 b) fx) = x e x c) ht) = sin t + ) Answer: f x) = x e x + x 9 e x Answer: h t) = cos t + ) t + ) t) d) rθ) = sin5θ + ) 8 cos5θ + π) Answer: r θ) = 5 cos5θ + ) + 4 sin5θ + π) e) fx) = x Answer: f x) = x ln ) f) fx) = secx) g) fx) = tan 4x)). Find the equation of the line tangent to the curve defined by at the point x =, y = π. Answer: f x) = secx) ln ) secx) tanx) Answer: f x) = tan 8 4x)) siny)) 3 cosy)) 3 + y π = x +4x) ) 3. A graph of fx) is given. In the axes below, make a reasonably accurate sketch of f x). fx) f x)

2 4. Consider the graph of fx) shown here: Make a reasonably accurate sketch of f x). Where is fx) not differentiable? 5. Find the tangent line and the normal line to the curve given by the graph of the function fx) = x 3 + x x at the point, ). 6. A radioactive element has a half-life of 53 years. Suppose that a sample initially contains 5 grams. a) Give a formula for at), the number of grams of the radioactive element in the sample after t years. b) Find a formula for the rate of change of the amount of the radioactive element. What are the units of this rate of change? What happens to the rate of change as t increases? 7. A chemical reaction is given by A + B C. Suppose d[a] = e 3t. What is d[b] d[c]? What is? 8. Section 3.3, Exercise ) A stone is dropped into a lake, creating a circular ripple that travels outward at a speed of 6 cm/s. Find the rate at which the area within the circle is increasing after a) s, b) 3 s, and c) 5 s. What can you conclude? 9. For each of the following, find. Your answers will be in terms of x and y.) a) x 3 + x + ) cosy) + y 5 = b) x 3 + sinxy) + y =

3 . See answers given with the functions.). We will use implicit differentiation: Solve for to get = Solutions 3siny)) cosy) 3cosy)) siny)) + π = 4x. At the point x =,y = π we find 4x 3siny)) cosy) + 3cosy)) siny) + π. So the equation of the line is = 4 π = 4π. y π = 4π)x ) or y = 4πx 3π. 3. Here are the graphs of fx) and f x): fx) f x)

4 Key observations: When fx) has a horizontal tangent, f x) is zero. These occur near x = 4.8, x = 3.5 and x =. In between these points, the sign of f x) agrees with the sign of the slope of fx). As x increases to the right, it appears that fx) is decreasing to zero. The slope is negative, and increasing towards zero. This shows up in the sketch of f x) as the curve approaches zero from below as x increases. 4. Just to the right of, the slope of the graph appears to be. As x increases from, the slope decreases. Just to the left of, the slope appears to be -. As x decreases, the slope stay negative but increases towards zero. A sketch of f x) looks like: The function is not differentiable at x = ; the graph appears to have a corner there, which shows up in the derivative as a jump discontinuity. 5. In both cases, we need to find f ). This will give us the slope of the tangent line, and the negative reciprocal of the slope of the normal line. f x) = 3x + 4x, and f ) = 6, so that tangent line is given by y = 6)x ), or y = 6x 6; and the normal line is given by y = x ), or y = x)/ a) We need a function that cuts the amount in half each time t is increased by 53. As we have seen in class and in the text Example 3, p. 6), such a function is at) = a ) t h, where a is the initial amount and h is the half-life. Thus, our function is at) = 5 ) t 53 = 5) t 53. 4

5 b) The rate of change is a 5 ln t t) = 53 ) 53, with unit of grams/year. The factor of 5 comes from the coefficient in front of at); the factor of ln comes from differentiating the exponential function with base ; the factor /53 comes from the chain rule.) As t increases, the rate of change approaches zero. In other words, the rate at which the material decays is decreasing. 7. For every molecule of A used up, a molecule of C is produced. Therefore, the rate of change of the concentration of C is just the negative of the rate of change of the concentration of A, so d[c] = d[a] = e 3t. On the other hand, for each molecule of A used up, two molecules of B are also used up, so the rate of change of the concentration of B must be twice the rate of change of the concentration of A. Thus d[b] = d[a] = e 3t. 8. Let At) be the area of the circle at time t. Then At) = πr t), where rt) is the radius of the circular ripple at time t. We are told that the ripple travels at a speed of 6 cm/s away from the center, so rt) = 6t. Thus At) = π)6t) = 36πt. The rate of change of the area is A t) = 7πt, so we find A ) = 7π cm /s, A 3) = 6π cm /s, and A 5) = 36π cm /s. The area grows at an increasing rate. 9. a) We must use implicit differentiation. Take the x derivative of both sides of the equation to get 3x + x + ) siny) ) + cosy)) + 5y 4 =, and then solve for : = 3x cosy) 5y 4 x + ) siny). b) This problem also requires implicit differentiation: 3x + cosxy) x ) + y + y =. Solve for to get = 3x y cosxy) x cosxy) + y. 5