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September 22nd, 2010, 05:22 PM   #1
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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:
 Problem #1: Evaluate the limit: 1/b - 1/2 ------------ lim b - 2 b->2
Quote:
 Problem #2 The displacement S (in meters) of a particle moving in a straight line is given by s=2t^3 where t is measured in seconds. (a) Find the average velocity (include units) of the particle over the time interval [8, 10] . Average velocity over [8,10] is 744 m/s (this one is incorrect) . (b) Find the instantaneous velocity (include units) of the particle when t=8 . Instantaneous velocity when t=8 is 384m/s (this one is correct) .
Quote:
 Problem #3 Let f(x)=4x^2 . (a) Use the limit process to find the slope of the line tangent to the graph of f at x=3 . Slope at x= 3 : (b) Find an equation of the line tangent to the graph of f at x=3 . Tangent line: y =
Quote:
 Problem #4 Let f(x) = 5x^4 - 2. (a) Use the limit process to find the slope of the line tangent to the graph of f at x=4 . Slope at x=4: (b) Find an equation of the line tangent to the graph of f at x=4. Tangent line: y =
Quote:
 Problem #5 Let f(x)=4x^2 - 5x. (a) Use the limit process to find the slope of the line tangent to the graph of f at x=2. Slope at x=2: (b) Find an equation of the line tangent to the graph of f at x=2. Tangent line: y =

 September 22nd, 2010, 05:32 PM #2 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 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, 05:49 PM #3 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 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, 08:37 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 y-intercept, 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, 08:41 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, 08:49 PM #6 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,209 Thanks: 517 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 point-slope 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 point-slope 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 point-slope formula gives y = 11(x - 2) + 6 = 11x - 16
September 22nd, 2010, 08:51 PM   #7
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Re: Calc Help

Quote:
 Originally Posted by MarkFL [f(x + h) = f(x)]/h
I assume you meant [f(x + h) - f(x)]/h.

September 22nd, 2010, 08:54 PM   #8
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Re: Calc Help

Quote:
Originally Posted by johnny
Quote:
 Originally Posted by MarkFL [f(x + h) = f(x)]/h
I assume you meant [f(x + h) - f(x)]/h.
Why yes I did! Thank you for catching that. Damn keyboard with "=" right next to "-"

 September 22nd, 2010, 08:56 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, 09:02 PM   #10
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Re: Calc Help

Quote:
 Originally Posted by johnny You're welcome Good job solving first half of the problems, where I solved the other half of problems

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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