h(t)=cot t

Question

a.) [pi/4,3pi/4] b.) [pi/6,pi/2]

Find the average rate of change of the function over the give n interval and intervals. Please show step by step instructions.

Thanks

Answer 1

The average rate of change = ( y2 - y1 ) / ( x2 - x1 )

a.)

x1 = π / 4

x2 = 3 π / 4

y1 = cot ( π / 4 ) = 1

y2 = cot ( 3 π / 4 ) = - 1

( y2 - y1 ) / ( x2 - x1 ) =

( - 1 - 1 ) / ( 3 π / 4 - π / 4 ) =

- 2 / ( 2 π / 4 ) =

( - 2 / 1 ) / ( 2 π / 4 ) =

- 2 ∙ 4 / 1 ∙ 2 π =

- 4 / π

b.)

x1 = π / 6

x2 = π / 2


y1 = cot ( π / 6 ) = √3

y2 = cot ( π / 2 ) = 0


( y2 - y1 ) / ( x2 - x1 ) =

( 0 - √3 ) / ( π / 6 - π / 2 ) =

- √3 / ( π / 6 - 3 π / 6 ) =

- √3 / ( - 2 π / 6 ) =

( - √3 / 1 ) / ( - 2 π / 6 ) =

- √3 ∙ 6 / 1 ∙ ( - 2 π ) =

- 6 ∙ √3 / - 2 π =

- 2 ∙ 3 ∙ √3 / - 2 ∙ π =

3 √3 / π

Answer 2

Of course in this case you can write:

The average rate of change = ( h2 - h1 ) / ( t2 - t1 )

t1 = π / 4

t2 = 3 π / 4

h1 = cot ( π / 4 ) = 1

h2 = cot ( 3 π / 4 ) = - 1

( h2 - h1 ) / ( t2 - t1 ) =

( - 1 - 1 ) / ( 3 π / 4 - π / 4 ) =

- 2 / ( 2 π / 4 ) =

( - 2 / 1 ) / ( 2 π / 4 ) =

- 2 ∙ 4 / 1 ∙ 2 π =

- 2 ∙ 4 / 2 π =

- 4 / π

b.)

t1 = π / 6

t2 = π / 2

h1 = cot ( π / 6 ) = √3

h2 = cot ( π / 2 ) = 0

( h2 - h1 ) / ( t2 - t1 ) =

( 0 - √3 ) / ( π / 6 - π / 2 ) =

- √3 / ( π / 6 - 3 π / 6 ) =

- √3 / ( - 2 π / 6 ) =

( - √3 / 1 ) / ( - 2 π / 6 ) =

- √3 ∙ 6 / 1 ∙ ( - 2 π ) =

- 6 ∙ √3 / - 2 π =

- 2 ∙ 3 ∙ √3 / - 2 ∙ π =

3 √3 / π

Answer 3

Bosnian, part b should be negative. Believe you solved as y2-y1/x1-x2 instead of y2-y1/x2-x1which would flip it to positive. Otherwise correct