MATH 125: LAST LECTURE


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1 MATH 5: LAST LECTURE FALL 9. Differential Equations A ifferential equation is an equation involving an unknown function an it s erivatives. To solve a ifferential equation means to fin a function that satisfies the given equation. Question. Consier the ifferential equation: 4y (y ) = y. This is an equation involving a function y an it s erivatives. Show that the function y(x) = x + 4x solves this ifferential equation. Answer. To check this, we calculate: an y (x) = x + 4 Now check: y (x) = 4y(x) (y (x)) = ( 4x + 6x 8 ) 4(x + ) = 4 = y (x). This function is a solution of this ifferential equation. In general, there will be many solutions to a given ifferential equation. We often select a particular solution of interest by perscribing an initial value. This is often calle solving an initial value problem.
2 FALL 9 Question. Solve the following ifferential equation y x = sin(x) +. Fin the specific solution that satisfies: y(3) = 5. Answer. To solve the above ifferential equation, we fin an antierivative. y(x) = cos(x) + x + C. Note the constant of integration. To fin the specific solution satisfying y(3) = 5, we use this information to etermine a specific constant of integration. 5 = y(3) = cos(3) C C = cos(3). In this case, the specific solution that satisfies y(3) = 5 is given bu y(x) = cos(x) + x + cos(3). In general, this is how you solve an initial value problem. Here is another example. Question 3. Show that y(x) = xe x + is a solution of the initial value problem: y x = ( x)e x with y() =. Answer 3. First, check the initial conition: y() = e + =. This checks! Now calculate: y (x) = e x xe x = ( x)e x. This checks too. So this is a solution of the above ifferential equation with this specific initial conition.
3 MATH 5: LAST LECTURE 3 Question 4. A water balloon launche from the roof of a builing at t = an has vertical velocity v(t) = 3t + 4 in units of feet per secon. Here v > correspons to upwar motion. a) If the roof of the builing is 3ft above the groun, fin an expression for the height of the water balloon. b) What is the average velocity between t =.5 an t = 3 secons? c) A 6 foot tall person is staning on the groun. How fast is the water balloon falling when it strikes this person on the top of the hea? Answer 4. a) We know the velocity: v(t) = 3t + 4. Since v(t) = s (t), we nee only fin an antierivative. Clearly, s(t) = 3 t + 4t + C = 6t + 4t + C. We were tol that the roof of the builing is 3 feet above the groun. Since the balloon is launche from the roof, the position of the balloon at t = is 3; i.e. s() = 3. This etermines a specific constant of integration. 3 = s() = 6() + 4() + C C = 3. The answer to question a) is then s(t) = 6t + 4t + 3. b) Recall the efinition of average velocity: Clearly, The Average velocity over [.5,3] = v(t)t ( 3t + 4)t = ( 6t + 4t ) t=3 t=.5 = 48. Thus 3 The Average velocity over [.5,3] = v(t)t = 48 = 3 ft/sec c) First, we can easily etermine the time t at which the balloon will hit the person on the hea. This comes from solving s(t) = 6. 6t + 4t + 3 = 6 6t 4t 4 = 6(t 3) ( t + ) = The only positive solution is t = 3. To fin the velocity at this time we evaluate This answers part c). v(3) = 3(3) + 4 = 56ft/sec
4 4 FALL 9 Question 5. If a car goes from to 8 miles per hour in 6 secons with a constant acceleration, what is the acceleration? Answer 5. Well, we know that the acceleration is constant. Set it to be k. a(t) = k. Since a(t) = v (t), we can fin the velocity using antierivatives. v(t) = kt + C. We can etermine the specific constant of integration using the fact that the initial velocity if. = v() = k() + C C =. Thus, v(t) = kt. We also know that v(6) = 8. So, 8 = v(6) = k(6) k = 8 = 3.33 meters/secon. 6
5 MATH 5: LAST LECTURE 5. The Secon Part of the Funamental Theorem of Calculus There are actually two important parts of the funamental theorem of calculus. We have seen the first part. Briefly, this says the following: F (b) F (a) = b a F (t)t. In wors, one can use antierivatives to evaluate efinite integrals. Here is an example: Question. Evaluate the following efinite integral. π sin(t)t =? Answer. Using the fact that the antierivative of sin(t) is cos(t), the answer is easy. π sin(t)t = ( cos(t)) t=π t= = + =. Notice that in the above efinite integrals, the variable t appears. This variable is calle the variable of integration. Of course, the results of these efinite integrals o not epen on t (the variable of integration). Once we have an antierivative, we can answer many questions. π/4 or π/ or 3π/ sin(t)t = ( cos(t)) t=π/4 t= = + = In fact, for any x we can evaluate sin(t)t = ( cos(t)) t=π/ t= = + =.. sin(t)t = ( cos(t)) t=3π/ t= = + =. x sin(t)t. Of course, the value of this efinite integral epens on the x value we choose. Hence, this quantity is a function of x. We can enote this function by f, i.e., set f(x) = x sin(t)t. The secon part of the Funamental Theorem of Calculus iscusses functions of this form.
6 6 FALL 9 Funamental Theorem of Calculus (n Part) If f is a continuous function on an interval an a is any number in that interval, then the function F efine by F (x) = x a f(t)t is an antierivative of f. In this case, Here is an example. Question. Consier the function for any x in the interval F (x) = f(x). x sin(t) Si(x) = t. t Observe that Si is a function of x. It is not a function of t, the variable of integration! Using the Funamental Theorem of Calculus, calculate the erivative of: a) Si(x) b) f(x) = xsi(x) c) g(x) = Si(x ). Answer. a) To calculate the erivative here, we just use the Funamental Theorem (part ). sin(x) Si(x) =. x x b) Here we use the prouct rule: sin(x) f(x) = Si(x) + x Si(x) = Si(x) + x = Si(x) + sin(x). x x x c) Here we use the chain rule. This is Si(x) compose with x. x g(x) = x Si(x) x=x x x = sin(x ) x x = sin(x ). x Question 3. Calculate ln(t) t. x x Answer 3. First, note that this is an upsie own integral (as far as the Funamental Theorem is concerne). To calculate it s erivative, rewrite it. ln(t) t = x ( ) ln(t) t = ln(x) = ln. x x x x
7 MATH 5: LAST LECTURE 7 Question 4. Calculate t sin(t) cos(x ) x. Answer 4. The easiest way to ifferentiate this function is to recognize that it is a composition. Set Set Observe that an so t sin(t) f(t) = t cos(x ) x f (t) = cos(t ). g(t) = sin(t) g (t) = cos(t). sin(t) cos(x ) x = f(g(t)) cos(x ) x = t f(g(t)) = f (g(t)) g (t) = cos ( (sin(t)) ) cos(t). Here is a similar question. Question 5. Calculate Answer 5. Set Set f(x) = 3 x 3 e t t. x cos(x) e t t f (x) = x 3 x e t t = x g(x) = cos(x) g (x) = sin(x). x 3 e t t = e x. Observe that 3 e t t = f(g(x)) cos(x) an so 3 e t t = x cos(x) x f(g(x)) = f (g(x)) g (x) = e cos (x) sin(x).
8 8 FALL 9 3. Integration via Substitution This is one of the most useful techniques of integration. The basic iea is as follows: () Take an ugly integral. () Rewrite the ugly integral as something you know how to integrate. (3) Calculate that integral. (4) Rewrite your answer in terms of the original quantities. Here is an easy problem: Question. Fin the antierivative of x 3 x =? We know the answer... x 3 x = x4 4 + C Let s get this answer a ifferent way. Answer. Let s rewrite the integral as x 3 x = x x x Introuce a new variable u = x. Clearly, u x = x. Let s write this symbolically as u = x x. Now substitute in this new variable: x x x = u u = u u. Now, in some sense, the integral on the righthansie is easier to calculate. We get u u = 4 u + C. In terms of x, what we have foun is: x 3 x = x x x = u u = 4 u + C = 4 x4 + C when we substitute back the fact that u = x into our answer. This is integration by substitution.
9 MATH 5: LAST LECTURE 9 Here is a harer problem. Question. Calculate 3x cos(x 3 ) x Answer. Try a substitution. Let u = x 3 u x = 3x u = 3x x. By regrouping, it is easy to see that 3x cos(x 3 ) x = cos(x 3 ) 3x x = cos(u) u = sin(u) + C = sin(x 3 ) + C, an we are one! As always, we can check our answers: x sin(x3 ) = cos(x 3 ) x x3 = cos(x 3 )3x, as we wante. Technically, substitution is reaing the chainrule backwars. Recall: If f an g are ifferentiable functions, then x f(g(x)) = f (g(x)) g (x). Thus if we integrate: f (g(x))g (x) x = f(g(x)) x = f(g(x)) + C, x by the first part of the Funamental Theorem of Calculus. This looks too complicate, so we make a substitution. If we want to calculate: f (g(x))g (x) x, make a substitution: Let an then u = g(x) u = g (x) x f (g(x))g (x) x = f (u) u = f(u) + C = f(g(x)) + C as before. The only har part about substitution is recognizing what to substitute! As above, often we substitute the insiefunction.
10 FALL 9 Question 3. Calculate Answer 3. Try a substitution. Let te t + t u = t + u = t t u = t t. By regrouping, it is easy to see that te t + t = e t + t t = e u u = e u u = eu + C = et + + C, an we are one! As always, we can check our answers: t et + = + et t (t + ) = te t +, as we wante. Question 4. Calculate Answer 4. Try a substitution. Let x 3 x x u = x u = 4x 3 x 4 u = x3 x. By regrouping, it is easy to see that x 3 x4 u x x = + 5x 3 x = 4 u an so u u = 4 4 u3/ 3/ + C = 6 (x4 + 5) 3/ + C an we are one! As always, we can check our answers: x 6 (x4 + 5) 3/ = 6 3 (x4 + 5) / 4x 3 = x 3 x 4 + 5, as we wante. Question 5. Calculate Answer 5. Try a substitution. Let e cos(θ) sin(θ) θ u = cos(θ) u = sin(θ) θ u = sin(θ) θ. By substituting, it is easy to see that e cos(θ) sin(θ) θ = e u u = e u + C = e cos(θ) + C, an we are one! As always, we can check our answers: θ ecos(θ) = e cos(θ) ( sin(θ)) = e cos(θ) sin(θ), as we wante.
11 MATH 5: LAST LECTURE Question 6. Calculate Answer 6. Try a substitution. Let e t + e t t u = + e t u = e t t. By substituting, it is easy to see that e t + e t t = u u = ln( u ) + C = ln( + et ) + C, an we are one! As always, we can check our answers: t ln( + et ) = + e t et, as we wante. Question 7. Calculate Answer 7. Try a substitution. Let tan(θ) θ u = cos(θ) u = sin(θ) θ u = sin(θ) θ. By rewriting, it is easy to see that sin(θ) tan(θ) θ = cos(θ) θ = u = ln( u ) + C = ln( cos(θ) ) + C, u an we are one!
12 FALL 9 You can also make substitutions into efinite integrals: Question 8. Evaluate x e x x Answer 8. One way to answer this problem is to use the Funamental Theorem of Calculus. In this case, we first fin an antierivative. Try a substitution. Let u = x u = x x u = x x. In this case, we fin that x e x x = e x x x = e u u = e u u = eu + C = ex + C This shows that ex is an antierivative, an so ( ) x e x x = ex t= t= = (e4 ). Question 9. Evaluate Answer 9. Try a substitution. Let π/4 tan 3 (θ) cos (θ) θ u = tan(θ) u = sec (θ) θ u = In this case, we fin that tan 3 (θ) cos (θ) θ = an so π/4 Question. Evaluate Answer. Try a substitution. Let cos (θ) θ. u 3 u = 4 u4 + C = 4 tan4 (θ) + C tan 3 ( ) (θ) cos (θ) θ = 4 tan4 (θ) θ=π/4 θ= = x x u = 5 x u = x u = x. In this case, we fin that 5 x x = u = ln( u ) + C = ln( 5 x ) + C u an so 3 x = ( ln( 5 x )) x=3 x= = ln() + ln(4) = ln(). 5 x
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