September 22nd, 2010, 04:22 PM  #1  
Newbie Joined: Sep 2010 Posts: 6 Thanks: 0  Calc Help
Hey I have a couple of problems that I'm looking for the solutions to if someone could help me out ... All and any help would be appreciated Quote:
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September 22nd, 2010, 04:32 PM  #2 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: Calc Help
For the first one, you can rewrite the expression (1/b  1/2)/(b  2) multiply by 2b/2b (2  b)/[2b(b  2) Use (2  b) = (b  2) (b  2)/[2b(b  2)] = 1/(2b) Now the limit can be found by the substitution b = 2, thus L = 1/4 
September 22nd, 2010, 04:49 PM  #3 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: Calc Help
For the second one, the average velocity over the interval [8,10] is found by: [1/(10  8 )][S(10)  S(] = (1/2)(2·10³  2·8³) = 1000  512 = 488 m/s the instantaneous velocity at t = 8 is found by: S'( S'(t) = 6t² = v(t), thus v( = 6·8² = 384 m/s 
September 22nd, 2010, 07:37 PM  #4 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Re: Calc Help
For the third one, lim delta x>0 (4(x+delta x)^2  4x^2)/delta x = lim delta x>0 (4(x^2 + 2x*delta x + (delta x)^2)  4x^2)/delta x = lim delta x>0 (4x^2 + 8x*delta x + 4(delta x)^2  4x^2)/delta x = lim delta x>0 (8x*delta x + 4(delta x)^2)/delta x = lim delta x>0 (8x + 4*delta x) = 8x + 4*0 = 8x + 0 = 8x. Therefore the slope of the line tangent to the graph of f at x=3 is 8 * 3 = 24. The slope is 24. y = ax + b, where a is the slope and b is the yintercept, so we have y = 24x + b. f(3) = 4 * 3^2 = 4 * 9 = 36, therefore y = 36 when x = 3, so we plug in the numbers in the y = 24x + b equation, which gives us 36 = 24*3 + b, 36  24*3 = 36  72 = 36 = b. Therefore equation of line tangent to y = 4x^2 is y = 24x  36. 
September 22nd, 2010, 07:41 PM  #5 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Re: Calc Help
For the fourth and fifth one, solve it like I did for the third one.

September 22nd, 2010, 07:49 PM  #6 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: Calc Help
For problems 3  5, use the difference quotient: [f(x + h)  f(x)]/h in the limiting process as h?0. For problem (3) this gives [4(x + h)²  4x²]/h = [4(x² + 2xh + h²)  4x²]/h = (8xh + h²)/h = [h(8x + h)]/h = 8x + h Now, as h?0 this gives 8x. So, m(x) = 8x, then m(3) = 24 We have slope m = 24 and point [3,f(3)] = (3, 36) so use of the pointslope formula gives y = 24(x  3) + 36 = 24x  36 For problem (4) we have: [(5(x + h)^4  2)  (5x^4  2)]/h = [5(x^4 + 4x³h + 6x²h² + 4xh³ + h^4)  5x^4]/h = 5h( 4x³ + 6x²h + 4xh² + h³)/h = 5(4x³ + 6x²h + 4xh² + h³) As h?0, we have 20x³ So, m(x) = 20x³, thus m(4) = 20(4³) = 1280. We have slope m = 1280 and point [4,f(4)] = (4,127 and again use of the pointslope formula gives y = 1280(x  4) + 1278 = 1280x  3842 For problem (5) we have: [(4(x + h)²  5(x + h))  (4x²  5x)]/h = [(4(x² + 2xh + h²)  5(x + h))  (4x²  5x)]/h = (8xh + 4h²  5h)/h = 8x + 4h  5 As h?0 this gives 8x  5 So, m(2) = 8(2)  5 = 11 We have slope m = 11 and point [2,f(2)] = (2,6) and again use of the pointslope formula gives y = 11(x  2) + 6 = 11x  16 
September 22nd, 2010, 07:51 PM  #7  
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Re: Calc Help Quote:
 
September 22nd, 2010, 07:54 PM  #8  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: Calc Help Quote:
 
September 22nd, 2010, 07:56 PM  #9 
Senior Member Joined: Apr 2007 Posts: 2,140 Thanks: 0  Re: Calc Help
You're welcome :) Good job solving first half of the problems, where I solved the other half of problems :) 
September 22nd, 2010, 08:02 PM  #10  
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,155 Thanks: 466 Math Focus: Calculus/ODEs  Re: Calc Help Quote:
It is painful to find derivatives with the limiting process once you know a few rules about differentiation. But, it is necessary to understand differentiation.  

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