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Using Calculus with Physics
Why Use Calculus?
 For situations where a value is not constant
o vf= vo+ at, vf2= vo2+ 2ad, d= vot+ ½at2 only work if “a” is constant
o Fg = GMm/r2 only works if “r” is constant
15
For Example: Using the graph at right…..
o Slope = acceleration… can only find slope of a straight line
(what if slope is not constant?)
o Area = displacement area is not a regular polygon
(how do you find the area for an irregular shape?)
Calculus terms and symbols:
 Slope = “Derivative” m = y2 – y1 =
x2 – x1
dx

10
v 5
(m)
0
0
1
2
T (s)
3
= dy = d’(x)
Area = “Integral”
length x width =
y x = y (dx)
Relationship Between Derivative and Integral:
 They are the opposite operations (division versus multiplication)
 One reverses the other:
 “Derivative” =

“Integral”
=
Examples of Calculus Use in Physics:
 DERIVATIVES: Slope for many physics graphs has MEANING….
 Any equation where there is division going on is in the form y is in slope (derivative)form.
v = d/t
d
a = v/t
v
t
W
t
d = dd = velocity
v = dv = accel.
Examples
of
Calculus
in Physics
t dt
tUsedt
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P = W/t
p = KE/v
V
t
W = dW = Power
(continued):
t
dt
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I = V/R
KE
R
v
V = dV = Current
R dR
KE = dKE = momentum
v
dv
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

INTEGRALS: Area for many physics graphs has MEANING….
Any equation where there is multiplication going on is in the form
y
d = vt
W = Fd
v = at
v
P = VI
a
V
x is in area (integral)form.
p = Ft
F
t
t
I
v (dt) = d
a (dt) = v
V (dI) = P
F
d
t
F (dd) = W
F (dt) = p
So How Exactly Does Calculus Work for me in physics?
DERIVATIVES – Use these for finding INSTANTANEOUS SLOPES



Suppose the position of an object is given by the equation d = do + vt. A little rearranging gives the
equation d = vt + do , which can readily be seen as a linear function y = mx+b with slope = velocity.
Slope = Derivative = velocity for this function (“Finding the slope” is the same as “taking the
derivative”)
Slope “m” = y2 – y1 =
= dy = d’(x)
20
x2 – x1
dx
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

Given that: x1 = t
x2
and y1 = kt+do and that
and y2 = k(t+
do …
10
d
(m)
The derivative (slope) of this line would be:
5
0
“m” = y2 – y1 = d2 – d1 = v2t2 – v1t1
x2 – x1 t2 – t1
t2 – t1


[v
(
o]
–t
– (vt+do) = vt + v
(
–t
0 +do –vt
1
–d2o = 3v
T (s)
=4 v
To summarize, the derivative (slope) of d = vt + do is v. (d’ of vt + do is v)
For the graph above, the equation would be: d = 2.5t + 5, and the slope (derivative) would be the
velocity (v) of 2.5 m/s.
Another way to get the same result…
d’(vt +do) = v = d(vt1 + doto)’/dt = 1vt1+ 0 doto = 1vt(1-1) + 0 = 1vt0 = v
Or, using our example graph equation…
d’(2.5t + 5) = d’(2.5t1 + 5t0) = 1(2.5t(1-1)) + 0(5t(0-1)) = 2.5t0 + 0 = 2.5 m/s
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Some other examples of finding the derivative (finding the instantaneous slope):





The derivative of 9t + 3 = 9
The derivative of 3t2+ 4t +5 = 6t +4
The derivative of 7t3+ 8t2 + 2t + 4 = 21t2 + 16t +2
The second derivative of 7t3+ 8t2 + 2t + 4 would be the derivative of its derivative, so the derivative of
21t2 + 16t +2 = 42t + 16
The derivative of 5t5/2 + 3t3/2 + 4t1/2 = 12.5t3/2 + 4.5t1/2 + 2t-1/2
INTEGRALS – Use these for finding AREAS for a certain x-INTERVAL



Remember that finding the Integral is the same as working the Derivative backwards.
To review our previous example, the derivative (slope) of d = vt + do is v
What would you have to do to work the derivative backward for this example?
o When you took the derivative, you reduced each exponent by 1, so now you must reverse that
trend by increasing each exponent by one:
Derivative (instantaneous slope or d’ or dd/dt ) of d = vt + do is vt(1-1) = vt0 = v
Integral (area of a v-t graph) is ( v dt) is vt(0+1) = vt1 + C (add back constant)
o For a more complicated integral, you must also consider the effect of reversing the trend on the
constants preceding the variable:
Derivative of d = 3t2 + 9t + 4 is 6t +9
6t1 + 9t0) dt is
Integral of 6t +9 =
? 6t(1+1) + ? 9t(0+1)
What number, multiplied by
6 would give the original prederivative constant of “3”?
Answer: 1/2
What number, multiplied by
9, would give the original
pre-derivative constant of
“9”?
Answer: 1
6t 1 + 9t0) dt = (½) 6t2+ (1) 9t + C or 3t2 + 9t + C
So the integral (area) of 6t + 9 is …
Some other examples of finding the integral (finding the area for a certain interval):





The integral of 9t + 3 = 4.5t2 + 3t
The integral of 3t2+ 4t +5 = t3 + 2t2 + 5t + C
The integral of 7t3+ 8t2 + 2t + 4 = 7/4 t4 + 8/3 t3 + t2 + 4t + C
The second integral of 7t3+ 8t2 + 2t + 4 would be the integral of its integral, so the integral of 7/4 t4 +
8/3 t3 + t2 + 4t + C = 7/20t5 + 8/12 t4 + 1/3 t3 + 2t2 + Ct + C ‘
The integral of 5t5/2 + 3t3/2 + 4t1/2 + C = 5/3.5 t7/2 + 3/2.5 t5/2 + 4/1.5 t3/2 + Ct + C’
Name __________________ Pd ____
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Some Practice: Finding Basic Derivatives and Integrals
Part I: Finding the DERIVATIVE (Instantaneous Slope)
STEP 1: Find the slope (take the derivative) of the following position and velocity equations:
STEP 2: Substitute the given time for the variable “t” and solve for the instantaneous velocity
or instantaneous acceleration.
Position (m)
1. d = 5t + 20
Velocity Equation
Instantaneous Velocity (m/s)
at t = 2 sec
This is easy…you know the slope!
2. d = 6t2- 3t -5
3. d = 9t3 + 7t2- 4t +7
4. d = 8
5. d = 4t3 – 8t
6. d = 7t5/2 + 5t3/2 – 6t1/2
7. d = 3t3/2 – 4t1/2 +6
Velocity (m/s)
Acceleration (m/s2)
8. v = 7t2 – 2t + 3
Instantaneous Acceleration (m/s2)
at t = 2 sec
9. v = 4t5 + 3t3 + 2t
10. v = 6t3 - 4
11. v = 5t + 7t2 – 3t3
12. v = 9t5/2 – 6t3/2 + 7t1/2
13. v = 5t-3 + 2t-2 – 3t-1
Some Practice (cont.):
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Part II: Finding the Integral (Area for a given time interval)
STEP 1: Find the area (take the integral) of the following velocity and acceleration
equations:
STEP 2: Substitute the given time interval for the variable “t” and solve for the total
displacement or total velocity. (consider that do is zero)
Velocity (m/s)
Displacement Equation
Total Displacement (m)
for t = 0-2 sec
Change in Velocity Equation
Total Change in Velocity (m/s)
for t = 0-2 sec
1. v = 5t + 20
2. v = 6t2- 3t -5
3. v = 9t3 + 7t2- 4t +7
4. v = 8
5. v = 4t3 – 8t
6. v = 7t5/2 + 5t3/2 – 6t1/2
7. v = 3t3/2 – 4t1/2 +6
Acceleration (m/s2)
8. a = 7t2 – 2t + 3
9. a = 4t5 + 3t3 + 2t
10. a = 6t3 - 4
11. a = 5t + 7t2 – 3t3
12. a = 9t5/2 – 6t3/2 + 7t1/2
13. v = 5t-3 + 2t-2 – 3t
Honors Calculus
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Application Problems
1. An object's position is described by the following polynomial for 0 to 10 s.
x = t3 − 15t2 + 54t
Where x is in meters, t is in seconds, and positive is forward. Determine …
a. the object's velocity as a function of time
b. the object's acceleration as a function of time
c. the object's maximum velocity
d. the object's minimum velocity
e. the object's average velocity
f. the object's average speed
2. An object's velocity, v, in meters per second is described by the following function of time, t, in
seconds for a substantial length of time …v = 4t (4 − t) + 8
a. Assuming the object is located at the origin (s = 0 m) when t = 0 s determine …
b. the object's position, s, as a function of time
c. the object's acceleration, a, as a function of time
d. the object's maximum velocity
e. if and when when the object stops
3. The following equations state displacement as a function of time. Derive the subsequent equations for
velocity and acceleration as functions of time. (The symbols A, f, j, k, x0, π andτ are all constants.)
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c. x = x0e−t/τ
b. x = A sin(2πft)
a. x = ⅙jt3
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4. A crude mathematical model of tunneling is represented by the equation … v = k/x
a. where v is the tunneling speed; x is the length of the tunnel; and k is a constant.
b. Determine the following quantities as a function of time …
i. tunneling speed
ii. tunneling acceleration
5. A simplified model of a car accelerating from rest along a straight path is given by the following
equation …v(t) = A(1 − e−bt), where v(t) is the instantaneous speed of the car in feet per second, t is the
time in seconds, and A and b are constants.
i. What is the speed of the car at t = 0 s?
ii. What is the limit of this function as t → ∞?
iii. Derive an equation x(t) for the instantaneous position of the car as a function of time. (Be
sure that your function has the value x = 0 m when t = 0 s.)
iv. What is the limit of this function as t → ∞?
v. Derive an equation a(t) for the instantaneous acceleration of the car as a function of time.
vi. What is the acceleration of the car at t = 0 s?
vii. What is the limit of this function as t → ∞?
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