AP Physics C
Essential Calculus
Examples of the calculus skills identified on the College Board AP Physics C Equation Sheet.
Differentiation - Power Rule
d n
x = nx n−1
dx
( )
Example #1
Given the function, y = y o + v o t +
1 2
gt
2
d
(y ).
dt
d 1 2gt 2
Find the first derivative,
( )
dy dy o d v o t
=
+
+
dt
dt
dt
dy
= 0 + v o + gt
dt
dy
= v o + gt
dt
v = v o + gt
(
)
dt
Find the second derivative,
d
(v ).
dt
( )
dv dv o d gt
=
+
dt
dt
dt
dv
= 0+g
dt
dv
=g
dt
a=g
Example #2
d n
du
u = nun−1
dx
dx
( )
Given v =
(
2ax = 2ax
)
1
2
1
= u 2 ; u = 2ax ;
du
= 2a
dx
1
1
d n
d ! 2 $ 1 −2
a
a
a
##u && = u 2a =
u =
=
=
dx
dx " % 2
2x
u
2ax
( )
( )
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Differentiation – Chain Rule
df df du
=
dx du dx
Example #3
Given the functions v x = at (1) and x =
1 2
dv
. From the equations
at (2). Find the acceleration, a =
2
dt
given it is obvious the acceleration is “a”.
Now write an equation for the velocity as a function of horizontal position. Solve equation (2) for time
and substitute into equation (1).
! 2x $
& = 2ax
v x = a ##
&
a
"
%
Again find the acceleration, a =
a=
dv dv dx dv
=
=
v
dt dx dt dx
dv 1
= 2ax
dx 2
(
a=
dv
. Since vx is not a function of time, apply the chain rule.
dt
)
−
1
2
2a =
a
2ax
! a $
dv
v =#
& 2ax = a
dx
" 2ax %
Differentiation – Trig Functions
d
sinx = cos x
dx
(
d
cos x = −sinx
dx
)
(
)
d
tanx = sec 2 x
dx
The derivatives of ALL trig functions starting with “c” are negative.
d
(sin u) = cos u du
dx
dx
d
(cos u) = − sin u du
dx
dx
Example
Find the first derivative
Let u = ωt + φ and
d
(A sin(ωt + φ))
dt
du
=ω
dt
d
(sin u) = A d (sin(Aω + φ)) = A cos(Aω + φ)(ω) = Aω cos(Aω + φ)
dt
dt
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(
)
Differentiation – The Product Rule
Always use the product rule; never the quotient rule.
d
(uv ) = u dv + v du
dx
dx
dx
Example
Differentiate f =
x
x2 + 1
(
)
Express f as product; f = x x 2 + 1
(
)
Let u = x and v = x 2 + 1
−
−
1
2
1
2
d
(f ) = d (uv ) = u dv + v du
dx
dx
dx
dx
3
1
d
(f ) = d (uv ) = x⎛⎜ − 1 ⎞⎟ x 2 + 1 − 2 (2x ) + x 2 + 1 − 2
dx
dx
⎝ 2 ⎠
(
d
(f ) = d (uv ) = −
dx
dx
d
(f ) = d (uv ) = −
dx
dx
d
(f ) = d (uv ) =
dx
dx
)
x2
(x
3
2
+1
)
1
+
) (x
2
(
2
1
)
+1 2
(x + 1)
(x + 1) (x + 1)
x2
2
3
2
2
+
2
3
2
1
(x
2
3
)
+1 2
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Differentiation – e
d x
e = ex
dx
( )
d u
du
e = eu
dx
dx
( )
Example
Differentiate e − kx Let u = −kx and
du
= −k
dx
d −kx
du
e
= eu
dx
dx
( )
d −kx
e
= e −kx (− k )
dx
( )
d −kx
e
= −ke −kx
dx
( )
Differentiation – ln
d
(ln x ) = 1
dx
x
d
(ln u) = 1 du
dx
u dx
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Integration – Power Rule
1
n
∫ x dx = n + 1 x
n +1
, n ≠ −1
Example #1
1
2
∫ kx dx = 2 + 1kx
2 +1
=
1 3
kx This is an indefinite integral.
3
Problems typically require limits of integration.
∫
3
2
kx 2 dx =
1 3 3 1 3 1 3
8
19
kx |2 = k3 − k 2 = 9k − k =
k
3
3
3
3
3
Example #2
1
n
∫ u du = n + 1u
2x
∫ (x
2
∫ (x
2
dx Let u = x 2 + 1 and du = 2xdx
3
dx =
2x
)
+1
, n ≠ −1
3
)
+1
n +1
−3 +1
−2
1
1
−1
x2 + 1
= − x2 + 1 =
2
2
− 3 +1
2
2 x +1
(
)
(
)
(
)
Example #3
∫ (x
x
2
3
)
+1
dx Let u = x 2 + 1 and du = 2xdx
From the equation du = xdx; need du = 2xdx; missing 2. Multiple by “1”.
2
1
2x
1 ⎛⎜
− 1 ⎞⎟
−1
dx
=
dx
=
=
3
3
2
∫
2 x2 + 1
2 ⎜⎝ 2 x 2 + 1 ⎟⎠ 4 x 2 + 1 2
x2 + 1 2
∫(
x
)
(
)
(
)
(
)
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Integration – Trig Functions
∫ sin x = − cos x
∫ cos x = sinx
Integration – e
∫e
x
= ex
u
u
∫ e du = e
Example #1
∫ − 2kxe
−kx 2
dx Let u = −kx 2 and du = −2kxdx
∫ − 2kxe
−kx 2
dx = ∫ eudu = e −kx
2
Example #2
2
−kx
2
∫ xe dx Let u = −kx and du = −2kxdx
From the equation du = xdx; need du = -2kxdx; missing -2k. Multiple by “1”.
∫ xe
−kx 2
2
2
− 2k
1
1
dx =
− 2kxe−kx dx = − e −kx
∫
− 2k
− 2k
2k
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Integration – ln
1
∫ x dx = ln x
Example #1
Integrate
∫
dv
where “v” represents velocity
v
dv
= ln v This is an indefinite integral.
v
∫
Problems typically require limits of integration.
⎛ v
dv
= ln v |vv if = ln v f − ln v i = ln⎜⎜ f
vi v
⎝ v i
ln x
Recall e = x
∫
vf
⎞
⎟⎟
⎠
du
= ln u
u
∫
Example #2
Integrate
kdq
∫ (1 + kq) where “q” represents charge
Let u = 1 + q and du = kdq
kdq
∫ (1 + kq) = ∫
du
= ln1 + q
u
Example #3
dq
∫ (k − q)
Let u = 1 − q and du = −dq
From the equation du = dq; need du = -dq; missing -1. Multiple by “1”.
dq
−1
∫ (k − q) − 1 = −1∫
(− 1)dq = − ln k − q
k−q
This is an indefinite integral.
Problems typically require limits of integration.
qf
dq
qf
0
∫ (k − q) = − ln k − q |
0
Recall e −ln x = e
ln
1
x
=
⎛ k − q ⎞
= − ln(k − qf ) − (− ln(k − 0 )) = −[ln(k − qf ) − ln(k )] = − ln⎜
⎟
⎝ k ⎠
1
x
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Differential Equations
B
d2 f x
( ) + Cf
dx 2
(x) = 0
B and C are constants
Rearrange so the second derivative term has no coefficients.
d2 f x
( )+Cf
dx 2
B
(x) = 0
Suppose the constants “B” and “C” equal 1.
d2 f x
( )+f
dx 2
(x) = 0
What function when differentiated twice and added to itself equals zero?
()
d2 f x
( ) = −sin x
()
dx
()
f x = sin x
2
OR
()
d2 f x
( ) = −cos x
()
dx
()
f x = cos x
2
Now let B≠1 and C ≠1
()
( )
Let f x = sin ωx and
d2 f x
( ) = −ω
dx
2
2
( )
sin ωx
Substitute the above equations into the general form of the differential equation.
d2 f x
( )+Cf
dx 2
B
(x) = 0
−ω2 sin ωx +
( )
C
sin ωx = 0
B
( )
Solve for ω
ω2 =
C
B
ω=
C
B
()
! C $
!
$
& OR f x = A sin # C x &
x
# B &
# B &
"
%
"
%
Final form of the solution. f x = sin #
()
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