apcalc exam review

how to evaluate limit

1. plug in value first, if you get number you’re finished

2. if you get 0/0 or infinity/infinity then simplify or use l’hopital’s rule → take derivative of both top and bottom

3. if you get a non-zero divided by 0 then its infinity or DNA

when is there a horizontal asymptote

when the limit approaches positive or negative infinity

when is a function continuous

if the limit exists (from left to right are the same) and the value of the function matches the limit.

also all differentiable equations are continuous

f(x) as x approaches a = f(a)

product rule differentiation

f’(x)g(x) + f(x)g’(x)

quotient rule

f’(x)g(x) - f(x)g’(x)/g(x)²

lowdehigh - highdelow/low²

derivative of lnx

1/x

derivative of e^x

e^x

derivative of log(base a)x

1/xlna

derivative of a^x

a^xlna

derivative of sinx

cosx

derivative of cosx

-sinx

derivative of tanx

sec²x

derivative of secx

secxtanx

derivative of cscx

-cscxcotx

derivative of inverse trig: write out

sin x→ x/sqrt (1-u²)

tanx →x/1+u²

sec x→x/absu * sqrt (u² -1)

cos x→ x/sqrt (1-u²)

cotx →x/1+u²

csc x→x/absu * sqrt (u² -1)

when is a function not differentiable

if it is not continuous or if the slopes from the left and right don’t match (corner) or the function is undefined at that point (vertical line)

implicit differentiation

use when u cant isolate y. take the derivative of each term. make sure to put dy/dx. then factor out dy/dx and solve for it

logarithmic differentiation

doing derivative by taking the ln of both sides

how to find the equation of a tangent line

1. find derivative. this will be the slope at that point.(find derivative equation then plug in that x coordinate to the derivative equation for the slope)

2. need a point. can be given or be solved for by plugging in the x value to the original equation

3. pput into y = y1 = m (x - x1) form. does not need simplification

describe 5 characteristics to a curve

1. increasing when first derivative is positive

2. decreasing when first derivative is negative

3. cc up when second derivative is positive (so first derivative is increasing)

4. cc down when second derivative is negative (so first derivative is decreasing)

5. points of inflection are when concavity changes

what is optimization

when you want to find the max or min of something

1. figure out what is being optimized

2. draw diagram and identify unknown variables

3. write what is being optimized in terms of the variables

4. use the constraint to write the equation in terms of one variable

5. take the derivative of the equation from 3

6. set equal to 0 and solve

7. check is max/min using the first derivative test or second derivative test ( make sure this is in the domain). for second derivative test if always negative cc down so is a max

8. use this one variable to solve for the other variable and therefore the two unknowns

how to solve a related rates problem

1. identify the rate given and the rate to find

2. find an equation that relates those variables then take the derivative of it with respect to time to solve for the rate it asks

from vid:

• make general diagram (w constant variables)

• make specific diagram (w changing variables)

• make an equation that related the variables in the general diagram

• take the derivative (make sure to put dX/dt for each bc of implicit differentiation)

• solve for the unknown using what is given in the equation

what are common related rates formulas to know

circle:

a = pir²

c = 2pir

rectangular prism

v = lwh

sa = 2lw + 2lh + 2wh

triangles: pythagorean theorem

a² + b² = c²

area: 1/2bh

cylinders:

v = pir²h

sa = 2pirh + 2pir²

spheres:

v = 4/3pir³

sa = 4pir²

right circular cone:

v = pir²h/3

if curve is increasing

left hand side = underestimate

right hand side = overestimate

if curve is decreasing

left hand side = overestimate

right hand side = underestimate

if curve is cc up

riemann sum is overestimate

if cc down

riemann sum is underestimate

the antiderivative with two limits that are functions of x

f(h(x)) h’(x) - f(g(x))g’(x)

to find the volume in general under a curve

take the antiderivative of the area * thickness (base times height or pir², etc)

volume for washer method

= pi * integral of R² - r²

• when when the strip does not sit on the rotating axis but when strip is perpendicular to it

use when given two equations & the area between the curves

volume for disk method

= pi * integral of r²

use when only given one equation

how to solve a net change problem

rate of it will always be the rate entering - rate leaving

• the integral of this rate is the net change (change in the amount of whatever it is measuring)

average value of an integral

1/b-a * integral of f(x) from a to b

all about motion

• v(t) = s’(t)

• a(t) = a’(t)

• moving right/up: when v(t) is positive

• moving left/down: when v(t) is negative

• speeding up: when v(t) and a(t) have the same signs

• slowing down: when v(t) and a(t) have opposite signs

• integral of velocity = change in positive

• integral of the absolute velocity = total distance traveled (split up into critical points and solve)

  • for displacement simply integrate between the intervals given

  • for total distance traveled set the v(t) equal to zero and solve. use this number to split up the integral into what you need to solve. add the absolute values of the integrals all together.

what is the condition & statement of the extreme value theorem

condition: f(x) is continuous over (a, b) *usually given in question

statement: f(x) has an absolute max and min value

there must be a max or min if continuous on a closed interval a-b

what is the condition & statement of the intermediate value theorem

condition: f(x) is continuous over (a, b)

statement: let L be a value such that f(a) < L < f(b), then there exists a c where f( c) = L and a < c < b

there must be a value f(c) =l between the closed intervals a-b

1. find f(a) and f(b) and calculate if l is between. if it is then it is equal to f( c). use this to then find c. make sure c is between a & b

what is the condition & statement for the mean value theorem

condition: f(x) is continuous over (a, b) and differentiable over (a, b)

statement: there exists a c such that f’( c) = f(b) - f(a)/b-a

• a < c < b

there must be a point w a tangent line slope that is the equal to the secant line connecting the endpoints (instantaneous rate of change at c = the average rate of change on interval a, b).

1. find average slope between f(a) and f(b)

2. take derivative

3. find the x value that makes the derivative equal to the average slope

4. solve for x

recirpcoal of tan

cot

reciprocal of sin

csc

reciprocal of cos

sec

derivative of cotx

-csc²x