7.4 Arc Length and Surface Area and other important concepts to know

Question 1

what do you need so that you can find arc length

Answer 1

–f is differentiable on a to b
–derivative of f is continuous on a to b

Question 2

equation for arc length

Answer 2

L = integral from a to b
square root of
1 + f prime^2

Question 3

what do you need so that you find surface area

Answer 3

f differentiable on a to b and its derivative f is continuous on a to b

Question 4

two equations for surface area

Answer 4

about the x-axis
SA = integral from a to b
2pi f square root of 1 + f prime^2

about the y-axis
SA = integral from a to b
2pi x square root of 1 + f prime^2

Question 5

how to solve arc length and surface area problems

Answer 5

just plug in
for surface area may need to revolve a graph first

Question 6

steps for completing the square

Answer 6

1. rearrange into
   ax2 +-bx = +- c
   —-factor a negative if a is negative
2. add (b/2)^2 to both sides
3. factor the left side and simplify the right side
   - diamond to factor
   -top is ac multiplied
   -bottom is add to b
4. make equal to 0 by +- the right side to left side

Question 7

the 3 differences of squares and the logic how to convert them

Answer 7

(a+b)^2 =
a^2 +2ab +b^2

(a-b)^2 =
a^2 -2ab +b^2

a^2 - b^2 =
(a+b)(a-b)

from standard to factored
just take a and b and make into the factored form

from factored to standard
you have a and b
find 2ab and convert into standard form

Question 8

logic for difference of squares with arc length and surface area

Answer 8

(a+b)^2 =
a^2 +2ab +b^2

(a-b)^2 =
a^2 -2ab +b^2

differ by 1
2ab and -2ab

it undoes itself so that ^2 and square root cancel out

(a-b)^2 + 1 = (a+b)^2

Question 9

how to factor a polynomial to a power > 2
- with 3 terms, powers of x of 3, 2, 1
-four terms, powers of x of 3, 2, 1, 0

Answer 9

3 terms
- factor GCF
-factor quadratic
-zero product property

4 terms
-split in half and group into two binomials being added
- take out the GCF and the insides should be the same
-if not rearrange og polynomial and try again
-if the insides the same, GCF plugs the other GCF times the binomial of the same insides is the factored form

Question 10

logic when you get the same og integral with integration by parts

Answer 10

a relationship

set the og equal to the other terms and the other og integral we got

combine those integrals and set equal to the other terms

solve and that is the answer

Question 11

the two special triangles for 6.4

Answer 11

for pi/3 or pi/6
A, O, H
1, 2, sqrt 3

for pi/4
A, O, H
1, 1, sqrt 2