Applications Of Derivatives

[Pages:4]Applications Of Derivatives

1

1. Find limx0 cos x2 6x.

Answer:

limx0 cos

1 x2

6x

e

1 2

62

2.

Find limx1

ln cos 6x6 x1 2

.

Answer:

limx1

ln cos 6x6 x1 2

18

3.

Find limx0

. sin x 2 sin2 x

x4

Answer:

limx0

sin x 2 sin2 x x4

1 3

4.

Find limx0

1 5

ln cos3x x2

.

Answer:

limx0

1 5

ln cos3x x2

9

10

5. Find limx? x2 x4 3 x4 .

Answer:

limx? x2

x4 3 x4

3

2

6.

Find limx?

1

4 x

x

Answer:

e4

7.

Find limx0

. sin x 2 sin2 x

x4

Answer:

limx0

sin x 2 sin2 x x4

1 3

8.

Find limx? x

61 x

1

.

Answer:

limx? x

61 x

1

ln 6

9.

Find

limx

1 12

ln sin6x

6x

1 2

.

Answer:

lim

x

1 12

ln sin6x

6x

1 2

0

10. Minimize f x 9 cot x 6 tan x for x 0, /2

Answer:

a cos2xbb cos2x cos2x 1cos2x

0. Thus cos x

1 5

10 . Thus you can find tan x, cot x. This yields 6 6 .

11. You want to find a solution to x 1 3 sin 10x

first iterate is at x1 2. Find the next iterate.

Answer:

1

310 cos10 1sin 20

0 using the Newton Raphson method. The

12. One car travels north at 40 miles per hour toward an intersection and the other travels east at 60

miles per hour away from the intersection. How fast is the distance between the two cars changing

when this distance equals 8 miles and the car heading north is at a distance of 3 miles from the

intersection?

Answer:

Let y be the distance to the intersection of the North bound car. Then

dy dt

40. Let x be the

distance to the intersection of the East bound car. Then

dx dt

60. The distance between the two

cars is D x2 y2 and so D x x y y. At the instant described,

x2y2

x2y2

D

15 2

55 15.

13. Two hallways, one 8 feet wide and the other 5 feet wide meet. It is desired to carry a ladder

horizontally arond the corner. What is the longest ladder which can be carried in this way?

Answer:

You need to maximize

5 sin8 cos cos sin

.

5 sin5 cos2 sin8 cos3 cos2 1cos2

.

Thus

you

need

tan3

8 5

and so

tan

3

8 5

. Then the minimum distance is the maximum length of the ladder which

3

is

35 24 .

14. A point x, y is picked on the ellipse x2 y2 1 which is in the first quadrant. Form the

16 9

rectangle which has this point and the point 0, 0 as ends of a diagonal of the rectangle. Out of all such rectangles, find the one which has maximum area.

Answer:

The area is A x

3 4

x

16 x2

.

Maximizing

this

is

the

same

as

maximizing

9 16

x2

16 x2

.

Take

the

derivative

of

this

to

get

9 4

x

8

x2

0, x 2 2 . Then the maximum area is 6.

15.

Find the point on the line

1 6

x

1 4

y

1

which

is

closest

to

0, 0

.

Answer:

You have

1 4

y

1

1 6

x

and

so

y

4

2 3

x.

Then

you

want

to

minimize

x2

4

2 3

x

2

. The

derivative is

26 9

x

16 3

0, x

24 13

.

Therefore,

the

closest

point

is

24 13

,

36 13

.

16. The surface area of a sphere of radius r equals 4r2 and the volume of the ball of radius r equals 4/3 r3. A balloon in the shape of a ball is being inflated at the rate of 3 cubic inches per minute. How fast is the surface area changing when the volume of the balloon equals 30 cubic inches?

Answer:

We have

3 r

V

A

as

a

relationship

between

area

and

volume.

Also

4/3

r 3

V and so

r

6 1 3

2

3

V 2

. Therefore, the relation is 3

36

2V2/3 A. Then differentiating, we get

3

36

2 V 1/3

2 3

V

A and so from the given information, A

3

36

2

30

1/3

2 3

3

2 5

32

33

35

2.

17. A hemispherical bowl of radius 8 inches is sitting on a table. Soup is being poured in at the

constant rate of 6 cubic inches per second. How fast is the level of soup rising when the radius of

the top surface of the soup equals 7 inches? The volume of soup at depth y will be shown later to

equal V y

8y2

y3 3

.

Answer: From the formula for the volume, V 6 16yy y2y and so when the radius is 7 and

y 15 , it follows 6 16 15 y 15 2y ,

y

6 16 15 15

.

18. A cylindrical can is to hold 80 cubic inches. The top and bottom of the can are constructed of a

material which costs 1 cents per square inch and the sides cost 3 cents per square inch

respectively. Find the minimum cost for such a can.

Answer:

You need r2h 80 and so h

80 r2

.

The

total

cost

is

then

C

r

2

r380 r

.

Then

C

r

4 r340

r2

0, Solution is : r

2 3

3 5 . Then the minimum cost is 24 3

35

2.

19.

The volume of a cone will be shown to be the

1 3

times the area of the base times the height later

on. Suppose you have a right triangle and the hypotenuse equals 6. You will revolve this triangle

around one of the other sides to form a right circular cone. What is the larges volume of such a

cone? Answer:

Let h be the height of this cone so that its radius is 36 h2 . Then the volume of the cone is

1 3

36

h2

h. How large can it be? Take the derivative 3h2 36 0, h 2

3 . Then the

maximum volume is 16 3 .

20. Find the maximum and minimum values if they exist of f x

ln 5x x

for x 0.

Answer:

1ln5x

x2

0, when x

1 5

e.

At

this

point,

the

value

of

the

function

is

5 e

.

This

function

doesn't

have a minimum and so this must be the maximum value.

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