Calculus I Practice Test Problems for Chapter 3 Page 1 of 9

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1 Calculus I Practice Test Problems for Chapter 3 Page of 9 This is a set of practice test problems for Chapter 3. This is in no wa an inclusive set of problems there can be other tpes of problems on the actual test. The solutions are what I woul accept on a test, but ou ma want to a more etail, an explain our steps with wors remember, I am intereste in the process ou use to solve problems! There will be five problems on the test. Most will involve more than one part. You will have 00 minutes to complete the test. You ma not use Mathematica or calculators on this test.. Fin the equation of the tangent line to the curve = x + cos x at the point (0,.. Fin the x coorinates on the curve x = 6x where the tangent line is horizontal. 3. Prove the Power Rule using logarithmic ifferentiation. The Power Rule: If n is an real number an f(x = x n, then f (x = nx n. 4. Prove the Sum Rule, [f(x + g(x] = x x f(x + x g(x]. 5. Fin /x if x + =. 6. Fin if = arctan x x. 7. Fin the secon erivative using implicit ifferentiation if x = 0. Note = x [ x ]. 8. Fin if = (sec x x + e sin x. 9. Use logarithmic ifferentiation to fin the erivative of the function = (x + 5 (x Given = (sin x x, show = (sin x x (ln(sin x + x cot x.. A man starts walking north at 4 ft/s from a point P. Five minutes later a woman starts walking south at 5 ft/s from a point 500ft ue east of P. At what rate are the people moving apart 5 mins after the woman starte walking?. A laer ft long rests against a vertical wall. If the bottom of the laer slies awa from the wall at a rate of ft/s, how fast is the top of the laer sliing own the wall when the bottom of the laer is 6 ft from the wall? 3. Show that [arctan x] = x + x.

2 Calculus I Practice Test Problems for Chapter 3 Page of 9 Solutions Problem. Statements: The slope of the tangent line is the erivative of the function. We want the equation of the tangent line, so our answer will look like 0 = m(x x 0. The point we are intereste in is (x 0, 0 = (0,, which has x = 0. We want to fin the erivative f (0 = m. We nee to efine f(x = x + cos x. Our answer will look like = f (0(x 0. f (x = [x + cos x] x = sin[x] f (0 = sin 0 = The equation of the tangent line to the curve at (0, is = f (0(x 0 = (x = x + Problem. Statements: The tangent line is horizontal means f (x = 0. However, x = 6x is an implicit function. We will nee to fin the x-coorinates where = /x = 0 (we cannot use f (x, since the function is implicit. x = 6x [ x x = 6x] implicitl ifferentiate [ x 3 ] + [ 3 ] = x x x [6x] 3x + [ 3 ] x = 6 [x] + 6x x x [] 3x + 3 x (3 6x x x = 6( + 6x x = 6 3x = 6 3x 3 6x Now, if /x = 0, we must have x = 0, or = x /. To get the x-coorinates for the points on the curve, substitute into the equation: x = 6x ( x x 3 3 ( x + = 6x x 3 + x6 8 = 3x 3 8x 3 + x 6 = 4x 3 x 6 = 6x 3 x 3 = 6

3 Calculus I Practice Test Problems for Chapter 3 Page 3 of 9 The onl real value solutions are x = 0 an x = 6 /3. Problem 3. Use logarithmic ifferentiation: = x n ln[ = x n ] ln[] = ln[x n ] ln[] = n ln[x] [ln[] x = n ln[x]] implicitl ifferentiate x ln[] = n x ln[x] ln[] x = n x chain rule x = n x x = n x = n xn x = nx n x [xn ] = nx n Problem 4. Let F (x = f(x + g(x. F (x = lim h 0 F (x + h F (x h = f(x + h + g(x + h f(x g(x lim h 0 h = f(x + h f(x g(x + h g(x lim + lim h 0 h h 0 h F (x = f (x + g (x [f(x + g(x] = x x f(x + x g(x (Leibniz notation Problem 5. Use implicit ifferentiation (ou can o this problem other was: x + = x [ x + = ] implicitl ifferentiate x [x ] + x [ ] = x [] x + [ ] x = 0 chain rule (an others x x = 0 x = x We have an implicit expression for the erivative, which is goo enough since the original equation was given implicitl.

4 Calculus I Practice Test Problems for Chapter 3 Page 4 of 9 Problem 6. If = arctan (x x, fin. = x = x [] = x [arctan (xx ] = [arctan (u], x u = u [arctan (u] u x = xx chain rule So no we nee two erivatives. If ou nee to, ou can work out the erivative of arctan u with respect to u using logarithmic ifferentiation, or if ou have it memorize just write it own. [arctan u] = u + u The other erivative requires logarithmic ifferentiation, so let w = x x : ln[w = x x ] ln w = x ln x [ln w = x ln x] x [ln w]w w x = [x ln x] x w w x = x [ln x] + ln x x x [x] w (x x = w x + ln x w x = xx ( + ln x Put it all back together, an substitute back u = x x : = + u xx ( + ln x = xx ( + ln x + x x Problem 7. Implicitl ifferentiate: x = 0 x [x3 + 3 = 0] x [x3 ] + x [3 ] = x [0] 3x + [3 ] x = 0 3x + 3 x = 0 x = x x = x [ ] x = x [x ] x x [ ] 4

5 Calculus I Practice Test Problems for Chapter 3 Page 5 of 9 (x x = [ ] x 4 ( x x x ( = 4 ( x x 4 + = 4 = x x4 5 Problem 8. Fin if = (sec x x + e sin x. x = x [] = x [(sec xx + e sin x ] = x [(sec xx ] + x [esin x ] The first erivative will require logarithmic ifferentiation; the secon erivative requires parts. w = (sec x x ln w = ln(sec x x ln w = x ln sec x [ln w] = [x ln sec x] x x w [ln w] w x w w x w w x w w x w x That was the har one! x [esin x ] = x [eu ] = x [ln sec x] + ln sec x x x [x] = x [ln u] + ln sec x( x u = x u [ln u] + ln sec x u x = sec x = x (sec x tan x + ln sec x u = (sec x x( x sec x tan x sec x = u [eu ] u x = e u (cos x = e sin x cos x Put it all back together: u = sin x + ln sec x x = x [(sec xx ] + x [esin x ] = (sec x x( x sec x tan x + ln sec x + e sin x cos x sec x Problem 9. Take the natural logarithm: = (x + 5 (x 4 3 6

6 Calculus I Practice Test Problems for Chapter 3 Page 6 of 9 ln = ln[(x + 5 (x ] = ln[(x + 5 ] + ln[(x ] ln = 5 ln(x ln(x 4 3 Implicitl ifferentiate: x ln = x [5 ln(x ln(x4 3] x = 5 x ln [ln(x + ] + 6 x [ln(x4 3] [x + ] x = 5 x + 6 x [x4 3] x + x 4 3 ( x = 5 x x3 x 4 3 ( 0 x = (x + 5 (x x + + 4x3 x 4 3 We backsubstitute for since the original equation was explicit, so our answer shoul be explicit as well. Problem 0. ifferentiation. Since we have a function of x raise to a power of x, we must o this erivative using logarithmic = (sin x x ln = ln[(sin x x ] ln = x ln sin x [ln ] = [x ln sin x] x x [ln ] x = [x] ln sin x + x [ln sin x] x x x = ( ln sin x + x x [sin x] sin x ( x = ln sin x + x cos x sin x x = (sin xx (ln sin x + x cot x Problem. We nee to raw a little sketch of the situation, an set up our notation. I ve rawn m sketch in Mathematica, but ou shoul raw our sketch b han.

7 Calculus I Practice Test Problems for Chapter 3 Page 7 of 9 Here, the man has move a istance of x from the point P, an the woman has move a istance from Q. The istance between the man an woman will be z. We are tol some information, an we can convert that information into mathematical equations using our notation. We are tol that the man walks with spee x t = 4 ft/s. We are tol that the woman walks with spee t = 5 ft/s. What is unknown is the rate at which the are moving apart, which is the rate of change of the istance between them, which is z t in our notation. To get the relation between x, an z, I am going to reraw m iagram. From the last iagram, we get the relation: z = (x What are x, an z after the woman has been walking for 5 minutes? = 5 ft/s = 300 ft/min so in 5 minutes the woman has walke = = 4500 ft. t x = 4 ft/s = 40 ft/min in 0 minutes the man has walke x = 0 40 = 4800 ft. t The istance between them after the woman has been walking 5 minutes is z = (x = ( = ft.

8 Calculus I Practice Test Problems for Chapter 3 Page 8 of 9 To introuce the rates of change, we must implicitl ifferentiate the relation with respect to time: z = (x t [z = (x ] t [z ] = t [(x + ] z z ( x = (x + t t + t z = x + ( x t z t + t The rate of change of istance between them after the woman has been walking for 5 minutes is z = t 00 ( = ft/min. Problem. We are given the rate of change of the istance the bottom of the laer is from the wall. What is unknown is the rate of change of the istance of the top of the laer from the groun. All our rates of change are with respect to time. Diagram/Notation: x is the istance of the base of the laer from the wall. is the istance of the top of the laer from the groun. z is the length of the laer. This is given as ft. x is given as ft/s. t is the unknown. t z is zero since the laer oes not change length. t From the iagram, we get the relation z = x + Implicitl ifferentiate with respect to time: t [z = x + ] z z = x x t t = + t ( t z z t xx t

9 Calculus I Practice Test Problems for Chapter 3 Page 9 of 9 Now we substitute in our values for when the bottom of the laer is 6 ft from the wall: z = ft x = 6 ft = z x = 6 = 6 3 ft The rates of change x/t an z/t were foun above. t = 6 ((0 (6( = 3 3 The top of the laer is sliing own (that s what the minus sign means! the wall at a rate of / 3 ft/s when the bottom of the laer is 6 ft from the wall. Problem 3. This is one with implicit ifferentiation. = arctan x x = tan Now implicitl ifferentiate x = tan [x x = tan ] x [x] = [tan ] x = [tan ] x = sec x x = sec = cos chain rule Now we have to get this back into an explicit function of x, since that is what we were given. Use SOH CAH TOA rules an a iagram of the situation to help us: x = tan = x = opposite ajacent + x cos = Therefore, x ajacent hpotenuse = + x, cos = + x. x = [arctan x] = x + x