Mathamatics,

br Show all work ( calculus 1

1 ) Show that the given families of curves are orthogonal trajectories of each other . Sketch both families of curves on the same axes

x2 y2 r2 , ax by 0 The two equations are orthogonal trajectories of each other (black circles for x2 y2 r2 , and the red lines are the family of ax by 0 You can see that any pair can be rotated along the axis with no change of shape

2 ) differentiate the function

a ) f (x x (1 /2 ) ln x [banner_entry_middle]

Using the derivative of products d (u v udv vdu

We let u x (1 /2 ) and v lnx

b ) y ln (x4Sin2x

let u x4Sin2x so that y becomes y ln (u ) and applying the derivative of product for d (u

3 ) Find y` and y

y x ln x

Using the derivative of products d (u v udv vdu

We let u x and v lnx

Solving for the 1st derivative y

Solving for the 2nd derivative y ‘ from y

4 ) Find an equation of the tangent line to the curve at the given point .y ln ln x (e , 0

Solving for the slope of the equation at any point m

we get the derivative using d (lnu (1 /u )du where u lnx

m y (x The slope of the tangent line mt is

mt m

Then we evaluate the value of the slope at x e

We get mt 1 /e

Using the point slope form y m (x-x1 y1 we get the equation of the tangent line

y mt (x-x1 y1 where x1 e and y1 0 we get the final answer

y (1 /e (x – e

5 ) Find the first and second derivatives of the function

y cos

The 1st derivative

y -sin

The 2nd derivative

y -cos

6 ) Find y `

y (2x 3 )1 /2

Applying dun n un-1 where u 2x 3

y (1 /2 (2x 3 )-1 /2 (2

y (-1 /2 (2x 3 )-3 /2 (2

y – (-3 /2 (2x 3 )-5 /2 (2

y 3 (2x 3 )-5 /2

7 ) If a snowball melts so that its surface area decreases at a rate of 1cm^2 /min , find the rate at which the diameter decreases when the diameter is 10cm

Since the equation of Surface Area (S ) as a function of diameter (d ) is

S d2

We get the derivative of both sides with respect to dt

Simplifying the equation by using rS for the rate of change of surface and using the given We can solve for the rate of change of diameter (negative meaning decrease

8 ) Find the critical numbers of the function

s (t 3t4 4t3 – 6t2

The critical numbers are found by getting the derivative and equating this to zero

s` (t 12t3 12t2-12t

t3 t2-t 0

t (t2 t-1 0

The critical numbers are

t0 0

9 ) Find the absolute max and absolute min values of f on the given interval

Solution : Get the derivative , equate to zero , solve for x , then get f (x )a ) f (x 3×2 – 12x 5 (0 ,3 0 6x -12

x 2

f (2 3 4 – 12 2 5 -7

b ) f (x 2×3 – 3×2 – 12x 1 ( -2 , 3

0 6 x2 – 6x – 12

0 x2 -x – 2

0 (x-2 (x 1

x1 2

x2 -1

f (x1 2 8-3 4-12 2 1

f (x1 16 -12 3 1

f (x1 -19

f (x2 2 (-1 )-3 (1 12 1

f (x2 -2-3 12 1 8 c ) f (x ( x2 – 1 )3 (-1 , 2

0 3 (x2-1 )2 (2x

0 6x (x2-1 )2

x1 0

x2 1

x3 -1

f (x1 1

f (x2 0

f (x3 0

d ) f (x x (x2 1 ( 0 , 2

f (x x (x2 1 )-1

0 – x (x2 1 )-2 (2x (x2 1 )-1

0 -2×2 (x2 1 )-2 (x2 1 )-1

0 -2×2 (x2 1 )-1 1

0 -2×2 (x2 1

0 -x2 1

x (-1 )1 /2 imaginary

f (x imaginary

d ) f (x ( ln x /x (1 ,3

0 – (lnx )x-2 x-1 x-1

0 1 – ln x

x e

f (x 1 /e

10 ) Find the most general antiderivative of the function ( check your answer by differentiation

Solution by integration . C denotes a constant

a ) f (x 10 /x9

f (x 10 x-9

F (x (-10 /8 )x-8 C b ) f (x 6 (x )1 /2 – (x )1 /6

F (x 6 (2 /3 )x3 /2 – (6 /7 )x7 /6 C

11 ) If 1200 cm2 of material is available to make a box with a square base and an open top , find the largest possible volume of the box

Solution

Let x be the width of the square box and y the height so the of open top considering 5 sides

1200 x2 4xy

y (x2-1200 /4x

y – (x2-1200 (4x )-2 (4x )-1 (2x

y – (x2-1200 8×2

y 7 x2 1200

0 7 x2 1200

x 1200 /7

x 171 .43 cm

y 41 .11 cm

largest volumen v

v x x y

v 1208150 .75 cm3

12 ) Write the composite function in the form f (g (x Identify the inner function u g (x ) and the outer function y f (u Then find the derivative dy /dx

y (4 3x )1 /2

let u 4 3x

y u1 /2

dy (1 /2 u-1 /2du

dy (1 /2 (4 3x ) -1 /2 (3dx

dy /dx (3 /2 (4 3x ) -1 /2

13 ) Find the derivative of the function

a ) f (t (1 tan t )1 /3

Solution

Dtf (t (1 /3 (1 tan t )-2 /3 (sec2t b ) y tan2 (3

Solution

dy /d 2tan (3 (3

dy /d 6tan (3

14 ) Find the most general antiderivative of the function ( check your answer by differentiation

a ) f (x x20 4×10 8

Solution

Axf (x (1 /21 ) x21 (4 /11 )x11 8x C

b ) f (x 2x 3×1 .7

Solution

Axf (x (2 /2 )x2 (3 /2 .7 )x2 .7 C

Axf (x x2 (3 /2 .7 )x2 .7 C

c ) f (x (x3 )1 /4 (x4 )1 /3

Solution

f (x x3 /4 x4 /3

Axf (x (4 /7 ) x7 /4 (3 /7 )x7 /3 C

d ) f (u u^4 3 (u )^1 /2 /u^2

15 ) Find f

f ` (x 2 – 12x , f (0 9 , f (2 15

Solution

1st Antiderivative of f (x

f (x 2x – (12 /2 )x2 C

f (x 2x – x2 C

2nd Antiderivative

f (x (2 /2 ) x2 – (1 /3 ) x3 Cx C2

f (x x2 – (1 /3 ) x3 Cx C2

3rd Antiderivative

f (x (1 /3 )x3 – (1 /12 ) x4 (C /2 )x2 C2x C3 let (C /2 C1

f (x (1 /3 )x3 – (1 /12 ) x4 C1x2 C2x C3

f (0 9 C3

f (2 (1 /3 )23 – (1 /12 ) 24 C1 22 C2 2 9 15

15 (8 /3 ) – (16 /12 4 C1 2 C2

No Solution : requires additional given f (x ) to solve

16 ) Given that the graph of f passes through the point (1 ,6 ) and that the slope of its tangent line at ( x , f (x ) is 2x 1 , find f (2

Solution

The slope is the 1st derivative

f (x 2x 1

1st Antiderivative

f (x x2 x C

Using the intersection to solve for C

6 f (1 1 1 C

C 4

We get the final equation f (x

f (x x2 x 4

So that

f (2 4 2 4

f (2 10

17 ) Find the differential of the function

a ) y cos (x

dy -sin (x (dx

dy – (sin (x )dx

b ) y x ln x

c ) y (1 t2 )1 /2

dy (1 /2 (1 t2 )-1 /2 (2tdt

dy t (1 t2 )-1 /2 dt

18 ) Use Part 2 of the Fundamental Theorem of Calculus to evaluate the integral , or explain why it does not exist

a ) The integration of 6 dx between 5 and -2

b ) The integration of (1 3y – y2 ) dy between 4 and 0

c ) The integration of x4 /5 dx between 1 and 0

d ) The integration of (3 / t4 )dt between 2 and 1

e ) The integration of cos )d ( between 2 ( and

19 ) Find a definition of `tangent` in a dictionary . Is it correct ? Other comments

From Wordweb

A straight line or plane that touches a curve or curved surface at a point but does not intersect it at that point

No this not entirely correct . It requires a mathematical such as a line with the same slope as the curve at the point of intersection

x

y … [banner_entry_footer]