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
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

Area = “Integral”
15
10
v 5
(m)
0
0
1
2
T (s)
3
= y = dy = d’(x)
x
dx
 Area =  length x width = 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” = y
x
 “Integral”
= y x
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.
x
v = d/t
d
a = v/t
v
t
d = dd = velocity
t dt
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P = W/t
W
t
v = dv = accel.
t dt
p = KE/v
V
t
W = dW = Power
t
dt
Mcps
I = V/R
KE
R
v
V = dV = Current
R dR
KE = dKE = momentum
v
dv
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Examples of Calculus Use in Physics (continued):


INTEGRALS: Area for many physics graphs has MEANING….
Any equation where there is multiplication going on is in the form yx is in area
(integral)form.
d = vt
v = at
v
P = VI
a
W = Fd
V
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”)
20
Slope “m” = y2 – y1 = y = dy = d’(x)
x2 – x1
x
dx
15
Given that: x1 = t and y1 = kt+do and that
x2 = t + t and y2 = k(t+t)+do …
The derivative (slope) of this line would be:
“m” = y2 – y1 = d2 – d1 = v2t2 – v1t1
x2 – x1 t2 – t1
t2 – t1


10
d
(m)
5
0
0
1
2
3
4
T (s)
[ v(t+t)+do] – (vt+do) = vt + vt +do –vt –do = vt = v
(t + t) – t
(t + t) – t
t
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.
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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
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
Integral of 6t +9 =
6t1
+ 9t0) dt
is
? 6t(1+1) +
What number, multiplied by
6 would give the original prederivative constant of “3”?
Answer: 1/2
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Mcps
? 9t(0+1)
What number, multiplied by
9, would give the original
pre-derivative constant of
“9”?
Answer: 1
2/6/2016
So
the integral (area) of 6t + 9 is … 6t 1 + 9t0) dt = (½) 6t2+ (1) 9t + C or 3t2 + 9t + C
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’
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Name __________________ Pd ____
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)
Instantaneous Acceleration
(m/s2) at t = 2 sec
8. v = 7t2 – 2t + 3
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
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Some Practice (cont.):
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
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