MATH 31B: MIDTERM 1 REVIEW. 1. Inverses. yx 3y = 1. x = 1 + 3y y 4( 1) + 32 = 1


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1 MATH 3B: MIDTERM REVIEW JOE HUGHES. Inverses. Let f() = 3. Find the inverse g() for f. Solution: Setting y = ( 3) and solving for gives and g() = +3. y 3y = = + 3y y. Let f() = Find a domain on which f is invertible. If g is the inverse of f, determine g ( 3). Solution: f () = , which is nonnegative when. f is increasing, hence onetoone, on [, ). After checking f() for some small values of, we find that f( ) = 3 = 3 so g( 3) =. By the inverse function theorem, g ( 3) = f (g( 3)) = f ( ) = 4( ) + 3 = 3 4 = 8. Find the derivative of +. Solution: Begin by writing. Eponential functions and logarithms + = e ln( +) = e ( +) ln() Now use the chain rule: d + d = d [ +) ln() d e( = e ( +) ln() + ] + ln()
2 JOE HUGHES. Evaluate d ln (). Solution: If u = ln(), then du = d. the integral is u du = u + C = ln() + C 3. Evaluate e e + d. Solution: Multiply the top and bottom by e to get e e e + d = e e d + e If u = e + e, then du = (e e ) d, so the integral is du u = ln(u) + C = ln(e + e ) + C Note that the term in the parentheses is always positive, so there is no need for absolute values. 3. Eponential growth and decay Two types of bacteria in a laboratory eperience eponential growth. Bacteria type A has an initial population of 000 and doubles every minute. Bacteria type B doubles every three minutes. If the two types of bacteria have the same population after 30 minutes, what was the initial population of bacteria B? Solution: Let A(t) and B(t) denote the populations of bacteria types A and B. Both are eperiencing eponential growth, so for constants k, k, C, C. A(t) = C e k t B(t) = C e k t To determine k, recall that the doubling time of A is equal to ln() k. Since the doubling time is one minute, this implies that k = ln(). Similarly, k = ln() 3. Net, for bacteria A we have And A(30) = B(30), so 000 = A(0) = C A(30) = 000e ln() 30 = ln() 30 = A(30) = B(30) = C e 3 = C 0 B(0) = C = 000 0
3 MATH 3B: MIDTERM REVIEW 3 4. y = k(y b). Find the solution to y + = 4y satisfying y(0) = 0. Solution: To solve this differential equation, begin by solving for y : The solution to this differential equation is y = (4y ) = y 6 = (y 3) y() = 3 + Ce To determine the constant C, plug in the initial condition: so C = 3. 0 = y(0) = 3 + C y() = 3 3e. A metal rod is heated to 00 degrees Celsius, then submerged in a large bucket of water whose ambient temperature is 0 degrees Celsius. After 5 minutes, the temperature of the rod is 0 degrees Celsius. The rod is then removed from the water. If the ambient temperature of the room is 5 degrees Celsius, what is the temperature of the rod after another 5 minutes? Solution: Begin by finding the formula for the cooling of the rod in the water. The equation for cooling is y(t) = T 0 + Ce kt = 0 + Ce kt and the constant C can be determined using the initial temperature: 00 = y(0) = 0 + C so C = 80. Now we want to determine the cooling constant k. To do this, use the fact that y(5) = 0: 0 = y(5) = e 5k so e 5k = =. ( 5k = ln = ln() ) so k = ln() 5. Now write out the equation for cooling once the rod is removed from the water: y(t) = T 0 + Ce kt = 5 + Ce ln() 5 t It s easiest to reset time to zero once the rod is removed from the water. The new initial temperature is the final temperature of the rod in the water, namely 0 degrees. 0 = y(0) = 5 + C
4 4 JOE HUGHES so C = 85. So the equation is To find the final temperature, plug in t = 5: y(t) = e ln() 5 t y(5) = e ln() = e ln() This is probably a fine answer, but you can also use the fact that e ln() = to simplify: in degrees Celsius. y(5) = = 35. Evaluate 0 e sin cos(ln( +)). 5. L Hopital s Rule Solution: Before using L Hopital s rule, always check that the it is in fact an indeterminate form. In this case, taking f() = e sin and g() = cos(ln( + )), we see that f() = 0 e0 = and g() = cos(ln()) = cos(0) = 0 L Hopital s rule does not apply, but we can simply plug in 0 to obtain. Evaluate π ( π ) tan. 0 Solution: Writing tan = sin cos gives e sin cos(ln( + )) = ( π ) tan π π ( π ) sin cos Let f() = ( π ) sin and g() = cos. Then f(π ) = 0 = g( π ) and f () = sin + ( π ) cos g () = sin which are both nonzero near π. L Hopital s rule can be applied, yielding π ( π ) sin cos 3. Show that ln() a = 0 for all a > 0. sin + ( π ) cos = π sin
5 Solution: Take f() = ln() and g() = a. hence we can apply L Hopital s rule to obtain since a > Evaluate ( + ). MATH 3B: MIDTERM REVIEW 5 ln() a Then f() = g(), aa Solution: Define L ( + ). Then ln(l) ln ( + ) a a = 0 ln( + ) This is an indeterminate form of the form 0 0, so by L Hopital s rule, L = e ln(l) = e = e. 5. Evaluate +. ln(l) = Solution: This is an indeterminate form of the form, so we can try to apply L Hopital s rule. The result is All we ve done is flipped the terms! In this problem, L Hopital s rule is valid but unhelpful. Instead, divide the top and bottom by, using = : =
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