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September 22nd, 2010, 04:22 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 =
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September 22nd, 2010, 04:32 PM   #2
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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
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September 22nd, 2010, 04:49 PM   #3
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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)(210 - 28) = 1000 - 512 = 488 m/s

the instantaneous velocity at t = 8 is found by:

S'(

S'(t) = 6t = v(t), thus v( = 68 = 384 m/s
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September 22nd, 2010, 07:37 PM   #4
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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.
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September 22nd, 2010, 07:41 PM   #5
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Re: Calc Help

For the fourth and fifth one, solve it like I did for the third one.
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September 22nd, 2010, 07:49 PM   #6
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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) - 4x]/h = [4(x + 2xh + h) - 4x]/h = (8xh + h)/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 + 4xh + 6xh + 4xh + h^4) - 5x^4]/h = 5h( 4x + 6xh + 4xh + h)/h =

5(4x + 6xh + 4xh + h)

As h?0, we have 20x

So, m(x) = 20x, thus m(4) = 20(4) = 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) - 5(x + h)) - (4x - 5x)]/h = [(4(x + 2xh + h) - 5(x + h)) - (4x - 5x)]/h = (8xh + 4h - 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
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September 22nd, 2010, 07:51 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.
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September 22nd, 2010, 07:54 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 "-"
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September 22nd, 2010, 07:56 PM   #9
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Re: Calc Help

You're welcome :)
Good job solving first half of the problems, where I solved the other half of problems :)
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September 22nd, 2010, 08:02 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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