Calculus Easy Slope of Curve Problem


by tensirk
Tags: calculus, curve, secant, slope
tensirk
tensirk is offline
#1
Feb7-11, 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=x3-8x; 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)/t-a?

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.
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eumyang
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#2
Feb7-11, 10:04 PM
HW Helper
P: 1,347
Maybe I'm dense, but have you learned how to take derivatives yet?
tensirk
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#3
Feb7-11, 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.

eumyang
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#4
Feb7-11, 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.
tensirk
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#5
Feb7-11, 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)=x3-x at x=1

Is this the same kind of problem I have above? The problem is not worked out in my book.
eumyang
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#6
Feb7-11, 10:26 PM
HW Helper
P: 1,347
See my previous post. (You posted while I edited, sorry!)
tensirk
tensirk is offline
#7
Feb7-11, 10:34 PM
P: 12
So, for the first one:

Vav= (1.01)^3-8(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!
eumyang
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#8
Feb7-11, 10:46 PM
HW Helper
P: 1,347
No, the first iteration is actually this line (bolded):
Quote Quote by eumyang View Post
Edit: Maybe do this?
Let a = 1, t = 1.1. Find Vav.
Then let a = 1 and t = 1.01. Find the new value of Vav.
...
So
[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]

Quote Quote by tensirk View Post
So, for the first one:

Vav= (1.01)^3-8(1.01) - 1 / 1.01 - 1 ? ....that comes out to -804.97 though.
But the bolded part is wrong. Check your math! f(1) ≠ 1.
tensirk
tensirk is offline
#9
Feb7-11, 10:46 PM
P: 12
I think I may have gotten it now.

Plug in y= x3-8x for both f(t) and f(a) and substitute the numbers of a and t into the equations and then divide by t-a.

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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