Calculus Easy Slope of Curve Problem

In summary, the conversation is about finding the slope of a curve at a given value of x. The equation given is y=x3-8x and the value of x is 1. The options provided are a. the slope is -3, b. the slope is 1, c. the slope is -5, and d. the slope is 3. The problem is not graded and the person is seeking guidance on how to solve it. A possible solution is to use the equation Vav= (f(t) - f(a))/(t-a) and plug in values for t and a to find the average slope of the secant line. By taking smaller and smaller values for t and a, the slope of the
  • #1
tensirk
12
0

Homework Statement


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.

Homework Equations


Would it be...
Vav= s(t)-s(a)/t-a?

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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  • #2
Maybe I'm dense, but have you learned how to take derivatives yet?
 
  • #3
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
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.
 
Last edited:
  • #5
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.
 
  • #6
See my previous post. (You posted while I edited, sorry!)
 
  • #7
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!
 
  • #8
No, the first iteration is actually this line (bolded):
eumyang said:
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]

tensirk said:
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.
 
  • #9
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. :)
 

1. What is the concept of slope in calculus?

The concept of slope in calculus refers to the rate of change of a curve at a certain point. It represents how steep or shallow the curve is at that point and is calculated by finding the derivative of the function at that point.

2. How do I find the slope of a curve in calculus?

To find the slope of a curve in calculus, you need to take the derivative of the function at the point of interest. This can be done by using the power rule, product rule, quotient rule, or chain rule depending on the function. Once you have the derivative, you can plug in the x-value of the point to get the slope.

3. What is the difference between average and instantaneous slope in calculus?

The average slope in calculus refers to the overall change in the curve over a given interval, while the instantaneous slope refers to the slope at a specific point on the curve. The instantaneous slope is more accurate as it takes into account the changes happening at that particular point.

4. Can I use calculus to find the slope of a straight line?

Yes, you can use calculus to find the slope of a straight line. However, since the slope of a straight line is constant, it can be calculated using the slope formula without the need for calculus.

5. How can I apply the concept of slope in real-life problems?

The concept of slope in calculus can be applied to various real-life problems, such as calculating the speed of an object, finding the rate of change of a quantity, or determining the optimal solution in optimization problems. It is a powerful tool for analyzing and predicting changes in a system.

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