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Calculus Easy Slope of Curve Problem 
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#1
Feb711, 10:00 PM

P: 12

1. The problem statement, all variables and given/known data
Find the slope of the curve for the given value of x. y=x^{3}8x; x=1 a. the slope is 3. b. the slope is 1. c. the slope is 5. d. the slope is 3. 2. Relevant equations Would it be... Vav= s(t)s(a)/ta? 3. The attempt at a solution I know this is a really simple problem, but I just can't figure it out and my book is no help at all. I know this problem deals with the slope of a secant line and I think I am supposed to make a table of values to plug into the equation of a secant line. Help is MUCH MUCH MUCH appreciated. This is not a problem that will be graded for homework. I have a Calculus test soon and have seemingly forgotten how to do the easiest problems (of course)...so a guided demonstration (or hints) of HOW to do this problem would make me extremely happy. 


#2
Feb711, 10:04 PM

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P: 1,347

Maybe I'm dense, but have you learned how to take derivatives yet?



#3
Feb711, 10:06 PM

P: 12

Nope, haven't even gotten that far...this is a problem from literally the first chapter of my book. It's definitely making me feel really dumb.



#4
Feb711, 10:14 PM

HW Helper
P: 1,347

Calculus Easy Slope of Curve Problem
Edit: Maybe do this?
[tex]f(x) = x^3  8x[/tex] [tex]Vav = \frac{f(t)  f(a)}{t  a}[/tex] Let a = 1, t = 1.1. Find Vav. Then let a = 1 and t = 1.01. Find the new value of Vav. Then let a = 1 and t = 1.001. Find the new value of Vav. Then let a = 1 and t = 1.0001. Find the new value of Vav. By this point, you can see what the slope of the curve will be at x = 1. 


#5
Feb711, 10:22 PM

P: 12

I've learned that, but it's in a later section than the problem I'm dealing with I think. I found a similar problem in my book where the directions state: "Slope of tangent lines: For the following functions, make a table of slopes of secant lines and make a conjecture about the slope of the tangent line at the indicated point."
The problem is: f(x)=x^{3}x at x=1 Is this the same kind of problem I have above? The problem is not worked out in my book. 


#6
Feb711, 10:26 PM

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P: 1,347

See my previous post. (You posted while I edited, sorry!)



#7
Feb711, 10:34 PM

P: 12

So, for the first one:
Vav= (1.01)^38(1.01)  1 / 1.01  1 ? ....that comes out to 804.97 though. Ahh, I'm sorry. I just am having a lot of trouble understanding this for some reason. Thanks so much for your help! 


#8
Feb711, 10:46 PM

HW Helper
P: 1,347

No, the first iteration is actually this line (bolded):
[tex]Vav = \frac{f(1.1)  f(1)}{1.1  1}[/tex] You did the 2nd iteration, [tex]Vav = \frac{f(1.01)  f(1)}{1.01  1}[/tex] 


#9
Feb711, 10:46 PM

P: 12

I think I may have gotten it now.
Plug in y= x^{3}8x for both f(t) and f(a) and substitute the numbers of a and t into the equations and then divide by ta. I got numbers arbitrarily close to 5, so I believe the answer is 5. Once again thanks SO much for all of your help. :) 


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