Answers for Calculus A Midterm Examination

Answers for Calculus A Midterm Examination

1.

Use implicit dierentiation to nd the tangent line at (2;2) of the graph of the function 2x^3;3y² = 4. Also nd the second derivative d²y=dx² at (2;2).

Sol : Using implicit dierentiation, we obtain

2x^3;3y² = 4; 6x^2;6ydydx ^{= 0}; 12x^;6(dy

dx^{)2 ;6}yd²y dx^{2 = 0}: At (2;2), 24^;12^dydx = 0, and 24^;6(^dydx)^2;12^ddx2y2 = 0. Then

dydx ^{= 2 and} d²y

dx^{2 = 0 at (2};2)

The tangent line at (2;2) is y^;2 = 2(x^;2), i.e. y= 2x^;2.

2.

For a ball with radius r meter, its surface area and volume are A = 4r² meter² and v = ⁴³r³ meter³, respectively.

(a) When r = 2 meter, what is the instantaneous rate of change of colume with respect to the radius? You should indicate the unit of your answer.

(b) Lethbe a very small positive number. When the radius increases from 2 meter to (2 +h) meter, use dierential to give an approximation on the increase of volume.

(c) Use geometric point of view to explain the relationship between the increase of volume derived in (b) to the surface area of the ball with radius 2 meter.

Sol : (a) The instantaneous rate of change of volume at r= 2 equals dVdr ^{= 4}r² = 16 (meter²)

(b) Since ^dVdr = 4r², for a smallh, V(r+h)^V(r) + ^dVdr
h = ⁴³r³+ 4hr². Forr = 2,
V =V(2 +h)^;V(2)^4h
2² = 16h.

(c) Since
V ^16hmeter³, and the surface area of the ball with radius 2 meter is 16 meter², the increase of volume
V is approximately the same as the surface area multiplies the increment of radius h.

3.

^{Set f(x) = x2sin}_0; ^{1x; x}_x^{6= 0}_{= 0}

(a) Find f⁰(0).

(b) When x⁶= 0, nd f⁰(x) =?

(c) Dosf⁰⁰(0) exist? If it exists, please nd its value. If not, give reason to support your argument.

Sol: (a) f⁰(0) = lim

h!0

f(h)^;f(0)

h ^{= limh!0} h²sin^1h

h ^{= limh!0}hsin 1h ^{= 0}: 1

(b) When x ⁶= 0, since x² and sin^1x are dierentiable, f(x) = x²sin^x1 is also dierentiable, and

f⁰(x) = d

dx⁽x²
sin 1x^{) = (} d

dxx^{2)
sin 1}x ⁺x²
( d

dx^{sin 1}x^{) = 2}xsin 1x^{;cos 1}x (c) f⁰⁰(0) = lim

h!0

f⁰(h)^;f⁰(0)

h ^{= limh!0(2sin 1}h ^; h^{1cos 1}h⁾:

If we make h tend to 0 by h = ^2n1 , the absoulute value of the limit tends to innity. So the limit does not exist.

4.

Sketch the graph of a rational function y = f(x) = x^2;2x+4

x^;2 . In the graph, you should discuss the concavity off, describe the rise and fall off, and nd asymptotes, local minimum and maximum if they exist.

Sol : y=f(x) = ^x2;2x+4x;2 =x+^x;24 ,f⁰(x) = 1^{; (x;2)4 2}, f⁰⁰(x) = ^{(x;2)8 3}. f⁰(x) = 0 i x= 0 or 4f(4) = 6,f(0) =^;2.

f⁰ >0 ifx >4 or x <0, the graph rises inx >4 or x <0.

f⁰ <0 if 0< x <2 or 2< x <4, the graph falls in 0< x <2, 2< x < 4.

f⁰⁰>0 if x >2, the graph is always concave up in x >2.

f⁰⁰<0 if x <2, the graph is always concave down in x <2.

The graph of

y=f(x) = ^x2;2x+4x;2

and its asymptotes:

−6 −4 −2 0 2 4 6 8 10

−20

−15

−10

−5 0 5 10 15 20 25

y = x O

x y

x = 2 (4,6)

(0,−2)

y=(x²−2x+4)/(x−2)

y=(x²−2x+4)/(x−2)

5.

Assume 0 ^ a ^ 1. Find the value of a such that ^R01jx^{2 ;} ax^jdx achieves its maximum.

Sol : 0^a^1,^jx^{2 ;}ax^j=



;x²+ax; 0^x ^a

x^2;ax; a^x^1 . Then

Z

1

0

jx^2;ax^jdx=

Z

a

0

(^;x²+ax)dx+

Z

1

a

(x^2;ax)dx 2

= (^;1

3x³_{+ 12}ax²)^ja0+(13x^3;1₂ax²)^j1a = 13a^3;1₂a_{+ 13 =}f(a):

The integral can achieve its maximum only at the end points 0, 1 or the point c²(0;1) for which f⁰(c) = 0.

f_{(0) = 13;}f_{(1) = 16}:

Since f⁰(c) =c^2;1₂;f⁰(c) = 0 i c² _{= 12}; orc= 1^p2;f( 1^p2) = 1 3^; 1

3^p2: So the integral achieves its maximum at a= 0.

6.

Find ^{dxd R2xx2}cos^ptdtwhen x >0.

Sol : Let F(x) = ^R0xcos^ptdt, since cos^pt is continuous, by fundamental theorem of calculus, we have F⁰(x) = cos^px, and for all a;b ^{2 R}, F(b)^;F(a) = ^Rabcos^ptdt. So when x >0,

dxd

Z

x 2

2x

cos^ptdt= d

dx^[F(x²)^;F(2x)] = F⁰(x²)
2x^;F⁰(2x)
2 = 2xcosx^;2cos^p2x:

7.

Evaluate the following integral:

(a)

Z sin(3t+ 2)

cos⁵(3t+ 2)dt; (b)

Z 1 x²

r

1^; 1 xdx:

Sol : (a) Letu= cos(3t+ 2), then du=^;3sin(3t+ 2)dt,

Z sin(3t+ 2)

cos⁵(3t+ 2)dt=^;1 3

Z du

u⁵ = 112u^;4+C _{= 112
cos}⁴₍₃¹t+ 2) +C (b) Letu= 1^;x1, then du= ^x12dx,

Z 1 x²

r

1^; 1 xdx⁼

Z

pudu_{= 23}u³² +C _{= 23(1}^{; 1}x^{)32 +}C

8.

Find the volume of the solid generated by revolving the shaded region (which is enclosed by x = 0 and x = 12(y^2;y³) as indicated in the following gure) about the x-axis.

−1 −0.5 0 0.5 1 1.5 2

−0.2 0 0.2 0.4 0.6 0.8 1 1.2

O

x y

x = 12 (y² − y³)

3

Sol : By shell method:

V =

Z

1

0

2y
12(y^2;y³)dy= 24

Z

1

0

(y^3;y⁴)dy

= 24(y⁴ 4 ^; y⁵

5 )^j10 = 24₍₁₄ ^{; 1}_{5) =} ²⁴ 20 = 6

5

9.

Suppose that y = ^x21+1, x^2R. Form a tangent line L=PQby picking two points P and Q in the graph of this funcrtion. Let m denote the slope of L. Show that

jm^{j }3^p4=8.

Proof : Give P = (a;^a21+1) and Q= (b;^b21+1), the slope of the line L is ^f(b);f(a)b;a By mean value theorem, there is a c betweena and b such that

f(b)^;f(a)

b^;a ⁼f⁰(c) = ^; 2c (c²+ 1)² Since d

dx^{(; 2}x

(x²+ 1)²) =^;2 1^;3x²

(x²+ 1)³ , f⁰⁰(x) attains its maximum and minimum at the point x=^p13. As x tends to innity,f⁰(x) tends to zero.

So the absolute maximun and minimum of the function f⁰(x) are f⁰(^; 1

p3) = 3^p3

8 ; f⁰( 1^p3) =^;3^p3

8 respectively.

So^jm^{j 3}

p3 8 :

4