MATH111-200630-PS02-Solutions

MATH111-200630-PS02-Solutions - Math 111 Problem Set 2...

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Unformatted text preview: Math 111 Problem Set 2 Solutions Edward Doolittle September 27, 2006 2. (a) Exponentiating both sides, ¥ ¡ ¨ ¥ (b) Taking the log base 2 of both sides, ¤ ¢ ¡ ¤ ¡ ¤ ¢  log2 23x © ¤ 4 log2 10 ¤ ¤ 3x Since there is no log2 button on my calculator I have to use the chain rule for logarithms to evaluate log2 10 ln 10 ln 2 ; with that, my calculator tells me x 0 2260. 3. (a) Write f x x ln x 1 2 . Then by the chain rule and product rule, ¤ ¥ ¤ f x x ln x ¢ 1 2 1  ¤ ¢ 2sin θ ¡ ¤ ¤ ¨ e ln 2 § ¡ ¢ ¢ sin θ d ln 2 sin θ dθ ln 2 cos θ ¨ § g θ d ln 2 e dθ sin θ ¨ § ¡ ¤ ¡ (b) Rewrite 2u in terms of the (natural) exponential function: 2u ¡ ¤ ¢ ¢  ¤ ¥ ¤ x ln x ln x © 1 2 1  ¢  ¥ ¤ x ln x ¢ ©   1 eln 2 u e ln 2 u . Then by the chain rule,   ¥ ¤ x ln x ¢ ©    1 ¢  ¢ 1 2 1 2 1 2 1 2 ¡ ¡ ¡ d x ln x dx d 2 x ln x dx 1 2 ln x x x x d ln x dx ¤  ¤ x ¢ ¢ ¢ ¤ 3x ¡ log2 10 4 1 log2 10 4 3 ¢ ¡ ¢ ¢ ¢ ¡ 4 log2 2 3x 4 log2 10 log2 10 ¡ ¢ ¢ © ¢ ¡ ¤ ¢  My calculator tells me that x 0 4002 to 4 decimal places.  ¤ ¥ x ¡ © ¢ ¥ 5x ¡ e 7 2 1 e 7 2 5 © ¡ ¥ 5x 2 ¡ § ¤ ¦ (b) ln e 8 1 ¡  ¤ £ ¤ ¡ ¤ ¢ ¡ ¤ ¢ ¡ ¤ ¡ ¤ 23 23 ln e 8. eln 5x 2 e e 7 7 ¥ £ 1. (a) 2 log10 3 ¢ log10 90 ¢ log10 90 log10 9 log10 90 log10 9 90 log10 10 ¡ ¡ ¤ ¡ ¡ log10 32 ¡ ¢ 1 1. 4. (a) Take the logarithm of both sides: ¥ ¤ 2 Then by implicit differentiation on the left side and the chain rule on the right side, ¤ ¤ © 4 ¡ Multiplying both sides by y and substituting what we know about y, 2x 1 2  Further simplification is possible, but not necessary unless the question asks for it. (b) Take the logarithm of both sides:  © ¢ ln h ¡ ln sin2 x cos4 x x2 1 3 ¤ ©  Differentiate the above equation (by implicit differentiation on the left side and the chain rule on the right side): © x2 1 Multiplying by h,  # © ¤ © ¢ h x 5. Differentiating both sides of the equation, then applying the chain rule,  ¡ Solving for y ,  2yy ¢ y y 2 ¤ y2  xex y 2 y 1 ex   ¢ ¡ 2 2 ¤  ¢ y  ¡  2y 1 2xex y 2y 1 ex2  ¡  ¤  ex 2 y2 2xex 2  ¡   2yex 2 y2 2xex 2 y2 y2 2  ¤ $ ©  ex ¢ 2 ¤ ©  ex ¡ ¢ ¡  d 2 y dx d 2y y dx 2yy  d x2 e dx 2  sin2 x cos4 x x2 1 3 2 cos x sin x 4 sin x cos x 6x x2 1 y2 d 2 x y2 dx y2 2x 2yy y2 y2 ¤ ¤  2 ¡ 3 ¢ © ¢ © ¡ ¤ ¢ ¡  h h 1 1 cos x 4 sin x sin x cosx 2 cos x 4 sinx 6x 2 sin x cos x x 1 1  ¤ 2 ln sin x 4 lncos x © 3 ln x2 1 2x 2 2x   ¢ © ¥ ¤ ¤ 1 x  © 2x3 4 1 2  ¢ © y y ¢ 24x2 2x3 1  1 24x2 2x3 1 1 ¢ © ¢ © ¡ ¡  y y 1 d 2x3 1 2x3 1 dx 24x2 1 2x3 1 2x 2 1 1 d x 2 x 1 dx ¤ ¤ © 4 ln 2x3 ¡ 1 1 ln x 2 ¢ ¤ ¢ ! ln y ln 1 x 1 "  © 2x3 4 ¢ ¢ 1 2 ¡ 1 1 ¡  ¡ 16 4 4 Integrating each term, 16 16 16 I ¡ x dx 4 6 dx 4 16 4 6x 0 0 0 0 ¢ © 7. (a) To find the inverse, solve for the independent variable t: so the inverse function is  ¤ £ ¥ Q ¤ ¥ 1 q a ln 1 q Q0 Q 1 q should be interpreted as the amount of time it takes for the camera to reach charge level q. 1 8. The slope of the line x 2y 5 is 1 2. The slope of any perpendicular line is the negative reciprocal, i.e., 2. We need to find a line tangent to y ex with slope 2. The line tangent to the curve at the point x ex has d x slope e ex , so we need to solve the equation ex 2: x ln 2 . A the point with the given x-value has dx coordinates ln 2 eln 2 ln 2 2 . The point-slope form of the tangent line is y 2 2 x ln 2 . 9. Using the hint and the rules for exponents, 3 2 3 x 0 x 0 0 3 3 2 5 ¤ 5  5 As x 0 , ln x 0 ¡ ¡ ∞, so ln x 2 ∞. Therefore limx xln x e∞  ¨ § 3 4 elimx ¡ 3 x 0 ln x 2 ¨ lim e ln x § 2 ¡ ¤ ¡ lim xln x 2 lim eln x ln x 2 ∞. ¤ 1 ¢ ¤ ¤ ¢ ¢ ¡ ¤ ¢ ¡ ¢ ¡ £ ¡ ¤ ¤ 1 ¢ ¡ ¢ ¢  Calculator says t 4 2 seconds approximately. ©  ¤ 6 ln 2 ¤ 3 ln 1 0 75 3 ln 1 4  ¤ £ ¢ ¢ ¤ £ ¤  ¢ ¡ ¡ ¡  ¥ Q 0 75Q0 3 ln 1 ¢ ¢ ¤ ¡ ¤  ¥ (b) By the previous result, we need to find Q ¢ 1 0 75Q0 : 0 75Q0 Q0 £ 1 ¢ ¡ ¤ ¡ ¤ ¢ £ £ ¢ ln 1 a ln 1 ¢ Q Q0 Q Q0 ln e t ¡ ¤  ¥ ¢ ¡ ¤ 1 ¢ e  ¥ 1 ¡ e t a  ¥ ¡ Q Q Q0 Q Q0 Q0 1 e t a t a ¤  ¥ ¢ ¡  To four decimal places, I 204 4766. t a t a ¤ ¤ 0  ¤ 0 © 128 192 0 ¡ 8 96 24 18 ln 2 9 ln 16 9 ln 4 ¢ ¢ 00 4 0 © © © 0 x2 2 0 ¡ 16 4 16 4 9 dx x 9 ln x   % © %  © ' I x dx x 6 © ©  % )¡ 3 x 2 16 9 x dx ¤ 6. Let I be the required result. First expand the integrand using the formula a © ' ( % &¡ b 2ab © © ¡ © ¢ 2 a2 b2: % ¡ ¡ ¤¨ § ¡ ¡ 1 ¤ ¢ ¢ ¡ ¢ 10. Here are two different solutions: (a) By formula 8 of section 7.4 and the continuity of the power function f x n ∞ n ∞ (b) By the continuity of the natural logarithm function, n ∞ n ∞ n ∞ n ∞ n ∞ by the result of problem 87 of section 7.4. Therefore ln lim 1 n ∞ and exponentiating both sides, n ∞ lim 1 ! 2 as required. 4 ¡ " a n © n ¡ " a n © 3 2 ! 2 £ n ∞ h 0 n a ea ¦ a lim 2 a lim a 1 ¡ ¡ ¤ ¡ ¤ ln 1 a n a n £ ln 1 h h © ¢ © ¢ £ Letting h a n the limit becomes a  £ 2 £ 2 £ 2 ! ln lim 1 ! ¡ 2 lim ln 1 2 lim n ln 1 a n lim a lim ¤ £ ¡ ¤ £ ¡ ¤ © ¢ ¡ " © ¡ " a n © n a n n ln 1 a n 1 n ln 1 a n a n © ¢ © ¢ £ where h a n. 3 2 3 h 0 ! 2  lim 1 ! 2 lim 2 1 lim 1 h lim 1 h 0 h ¡   ¤ ©  ¡ "  ¤ 1 h ¢ © ¢ ¡  " © !  ¡ " a n © n a n n a a a ¡ ¤ ¢ xa , a 1 h ea ¡ ...
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This note was uploaded on 01/12/2010 for the course MATH 111 taught by Professor Doolittle during the Fall '06 term at University of California, Berkeley.

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