**Unformatted text preview: **Math 1205 --Test 2
Instructor: 17 October 2007 NAME: You will be graded on how well and fully you support your work! Use only methods developed in class. 1.(10 pts) For the given graph, mark whether the following are true or false: a.) f'(x) exists when x= - 3. True b.) f '(2)=0. False c.) The derivative of f(x) exists when x= - 2. False d.) The derivative of f(x) has a right hand derivative when x= 2. False e.) f(x) has a left hand derivative when x=3. True
4 3 2 1 -4 -2 -1 2 4 2. I10 pts.M Using the limit definition, find the derivative of y = x+1+h h x+1+h h x+1 x+1+h + x+1+h + x+1 = x+1 h h x+1 = x + 1 for all x > -1. -> x+1 2 1 x+1 x+1+h + 3. I6 ptsM Let y = 1 4 x 4 + 2 x 3 - 5 x 2 + 7. Find y'''. y ' = x3 + 6 x2 - 10 x, y '' = 3 x2 + 12 x - 10, y ''' = 6 x + 12, 4. (10 pts) The motion of a particle is given by s(t). The graph of s(t) is given below. On the same axes graph the velocity v(t) given the fact that the maximum speed of the object is about .44.
1.0 0.8 2 3. I6 ptsM Let y = 1 4 x 4 + 2 x 3 - 5 x 2 + 7. Find y'''. y ' = x3 + 6 x2 - 10 x, y '' = 3 x2 + 12 x - 10, y ''' = 6 x + 12,
test2_final_solutions.nb 4. (10 pts) The motion of a particle is given by s(t). The graph of s(t) is given below. On the same axes graph the velocity v(t) given the fact that the maximum speed of the object is about .44.
1.0 0.8 0.6 0.4 0.2 -6 -4 -2 -0.2 -0.4 2 4 6 5. (6 pts each) variable : a.) g(t) = Find the derivative for each of the following with respect to the given independent t2 + 3 t + 1
I-12M b.) k(p) = sin3 (5p + 2) 3sin2 (5p+2)cos(5p+2)(5)=15sin2 (5p+2)cos(5p+2) It 2 + 3 t + 1M (2t+3) c.) P(q) = (3 q + 2)/I5 q + 1M 4 [I5 q + 1M (3)-4I5 q + 1M (5)(3q+2)] I5 q + 1M =
4 3 8 3 I5 q + 1M (3)-20I5 q + 1M (3q+2)]/I5 q + 1M
4 3 8 6. (6 pts each) In each of the following y is a function of x. a.) Let y = sin(pxy/2). Find dy/dx in terms of x and y and evaluate it at the point (1,1). y'=cos(pxy/2)(pxy)' or y'=cos(pxy/2)(p/2)(xy'+y) test2_final_solutions.nb 3 6. (6 pts each) In each of the following y is a function of x. a.) Let y = sin(pxy/2). Find dy/dx in terms of x and y and evaluate it at the point (1,1). y'=cos(pxy/2)(pxy)' or y'=cos(pxy/2)(p/2)(xy'+y) y'(1-cos(pxy/2)(p/2)x)=cos(pxy/2)(p/2)y y'=(p/2)y(cos(pxy/2))/(1-(p/2)xcos(pxy/2)) Plug in to get y'=0 at (1,1). b.) Let x 2 + y 2 = 1. Find d 2 y/dx 2 in terms of x and y. 2x+2yy'=0 gives y'=-x/y So y''={y(-1)-(-x)y'}/y 2 ={-y+x(-x/y)}/(y L2 ={-(y L2 -(x L2 /(y L3 =-1/(y L3 7. (6 pts each) Find a.) y=3x y'=log3 3x dy dx for each of the following: b.) y=sin-1 (log x) (Recall that in our notation log(x) is the same as ln(x).)
y'= [1/ I1 - Ilog x M^2 )]/x 8. (6 pts each) The following problems involve logarithmic differentiation. a.) Find
dy dx at x= 1 when y= Ix +1M Ix 2 +1M
2 x 5 +1 y'=y[(2/(x+1)) +( 1/(x 2 + 1))(2x) - ( 1/(x 5 + 1))(5x 4 )] When x=1, y=4, so y'=-2 b.) Find dy dx when y=x cos x log y=log x cos x , log y = cos x log x. y'/y = (cos x)/x - sin x log x) or y'=(cos x)/x - sin x log x)x cos x . y'=y[(2/(x+1)) +( 1/(x + 1))(2x) - ( 1/(x + 1))(5x )] When x=1, y=4, so y'=-2
4 test2_final_solutions.nb b.) Find dy dx when y=x cos x log y=log x cos x , log y = cos x log x. y'/y = (cos x)/x - sin x log x) or y'=(cos x)/x - sin x log x)x cos x . 9. (10 pts) A rocket moves from left to right along the curve y= x 2 . When the engines are shut off the rocket will go off along the tangent line at the point where it is at the time the engines are shut off. At what point on the curve should the engines be shut off so that the rocket will reach the point (6,32)? We denote a to be the answer. The slope of the tangent line is 2a and the line passes through the points (6,32) and (a,a 2 ). We have 32-a 2 =2a(6-a). Factoring, we find that a=4 or a=8, We choose a=4 since when a=8 we are already past the target. ...

View
Full Document

- Spring '08
- FBHinkelmann
- Math, Calculus, Derivative, Cos, Logarithm, dy dx