Using Calculus with Physics

Why Use Calculus?

  • For situations where a value is not constant
  • vf=vo+at, vf2= vo2+ 2ad, d=vot+½at2 only work if “a” is constant
  • Fg = GMm/r2 only works if “r” is constant

For Example: Using the graph at right…..

  • Slope = acceleration… can only find slope of a straight line

(what ifslope is not constant?)

  • 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 = y = dy = d’(x)

x2 – x1 x dx

  • Area = “Integral”  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 yis in slope(derivative)form.

x

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xis in area(integral)form.

So How Exactly Does Calculus Work for me in physics?

DERIVATIVES – Use these for finding INSTANTANEOUSSLOPES

  • 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+bwith slope = velocity.
  • Slope = Derivative = velocity for this function (“Finding the slope” is the same as “taking the derivative”)
  • Slope “m” =y2 – y1 = y = dy = d’(x)

x2 – x1 x dx

  • 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 [v(t+t)+do] – (vt+do) = vt + vt +do –vt –do = vt = v

x2 – x1 t2 – t1 t2 – t1 (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.

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 AREASfor 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?
  • 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 isvt(1-1)= vt0 = v

Integral(area of a v-t graph) is (v dt) is vt(0+1) = vt1 + C (add back constant)

  • 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) + ?9t(0+1)

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’

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) / Velocity Equation / Instantaneous Velocity (m/s)
at t = 2 sec
1. d = 5t + 20 / 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

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 velocityand 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
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) / Change in Velocity Equation / Total Change in Velocity (m/s) for t = 0-2 sec
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

H:\Calculus\Using Calculus with Physics.docMcps9/12/2008