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Particular solutions vs general solutions Position, velocity, acceleration problems
You should realize that an antiderivative of a product or a quotient CANNOT be handled in parts!  Just like we’ve seen with derivatives of products and quotients, we can’t derive things “in place”.  That also applies to products or quotients inside integrands.  They have to be rewritten to be able to use an existing rule.  Two other warnings:  Always include the “+ C” for the indefinite integrals (antiderivatives) and ALWAYS make sure the dx or d-whatever is in the integrand.  I look at those very critically!
Initial Conditions and Particular Solutions  You have seen that we get a family of solutions when we solve  y  =  ∫ f(x) dx since we always add a constant of integration, + C onto the end.  This means that the graphs of any two members of the family are VERTICAL translations of each other. y =  ∫ (3x 2  – 1)dx y = x 3  – x + C Through (2, 4)
y = x 3  – x + C is the general solution, and we want the particular solution that goes through the point (2, 4) 4 = 2 3  – 2 + C, so C = -2 The particular solution is y = x 3  – x - 2
Ex 7, p.253  Finding a particular solution Find the general solution of F’(x) = 1/x 2  .  Then find the particular solution that goes through the point (1, 2) Solution:  Find F(x) =  ∫  Rewrite F’(x)  Find general solution Plug in (1, 2) to find C. Write particular solution GO!
Ex. 8, p. 254  Solving a vertical motion problem ,[object Object],[object Object],[object Object],What do we know? s’(t) =  ∫-32dt  and s’(t) = -32t + C 1 .  But s’(0) = 64, so 64 = -32(0) + C 1   and C 1  is 64. s’(t) = -32t + 64 s(t) =  ∫( -32t + 64)dt = -16t 2  + 64t + C 2 . But s(0) = 80;  80 = -16(0) 2  + 64(0) + C 2  and C 2  is 80 s(t) = -16t 2  + 64t + 80  which is answer to part a.
[object Object],[object Object],[object Object],5 seconds
We can do the same kind of process to analyze linear motion problems (vertical or horizontal) that is the result of an acceleration due to some other force besides gravity. 4.1b p. 255/ ( 35-45 odd), 49, 51, 55-61 odd, 67-74 all, 77, 78

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Calc 4.1b

  • 1. Particular solutions vs general solutions Position, velocity, acceleration problems
  • 2. You should realize that an antiderivative of a product or a quotient CANNOT be handled in parts! Just like we’ve seen with derivatives of products and quotients, we can’t derive things “in place”. That also applies to products or quotients inside integrands. They have to be rewritten to be able to use an existing rule. Two other warnings: Always include the “+ C” for the indefinite integrals (antiderivatives) and ALWAYS make sure the dx or d-whatever is in the integrand. I look at those very critically!
  • 3. Initial Conditions and Particular Solutions You have seen that we get a family of solutions when we solve y = ∫ f(x) dx since we always add a constant of integration, + C onto the end. This means that the graphs of any two members of the family are VERTICAL translations of each other. y = ∫ (3x 2 – 1)dx y = x 3 – x + C Through (2, 4)
  • 4. y = x 3 – x + C is the general solution, and we want the particular solution that goes through the point (2, 4) 4 = 2 3 – 2 + C, so C = -2 The particular solution is y = x 3 – x - 2
  • 5. Ex 7, p.253 Finding a particular solution Find the general solution of F’(x) = 1/x 2 . Then find the particular solution that goes through the point (1, 2) Solution: Find F(x) = ∫ Rewrite F’(x) Find general solution Plug in (1, 2) to find C. Write particular solution GO!
  • 6.
  • 7.
  • 8. We can do the same kind of process to analyze linear motion problems (vertical or horizontal) that is the result of an acceleration due to some other force besides gravity. 4.1b p. 255/ ( 35-45 odd), 49, 51, 55-61 odd, 67-74 all, 77, 78