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(a) Find the slope of the tangent to the curve at the point where x = a.

by Pondera
Tags: curve, point, slope, tangent
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Pondera
#1
Feb8-09, 09:40 PM
P: 13
1. The problem statement, all variables and given/known data
Consider the following curve.
y = 4 + 4x2 - 2x3.

(a) Find the slope of the tangent to the curve at the point where x = a.

2. Relevant equations
m=y(a+h) - y(a)/h

3. The attempt at a solution
1. m=y(a+h) - y(a)/h = -a^3-a^2+3ah+h^2-4
2. 6a^2
3. -6a^2+8a

I know the derivative is equal to the slope, but I don't know how to work that into this specific equation and point.

Thank you for any and all help.
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Mark44
#2
Feb8-09, 10:03 PM
Mentor
P: 21,397
Do you know any differentiation formulas, or do you have to do this by using the definition of the derivative?

The equation you give as the relevant equation works for equations of lines, but not for any other equations.

If you have to use the deriviative definition, let's call your function f(x) = -2x^3 + 4x^2 + 4.

By the definition,
[tex]f'(x) = \lim_{h \rightarrow 0} \frac{f(x + h) - f(x)}{h}[/tex]

You skipped a lot of steps apparently in your attempt at a solution. Start by evaluating f(x + h), and subtract f(x) from that. Then divide each term in f(x + h) - f(x) by h, and finally, take the limit as h approaches zero.

That will be f'(x). To get f'(a), substitute a for x in your derivative function.
Pondera
#3
Feb8-09, 10:23 PM
P: 13
This is the only formula we've learned so far, though I'm sure we'll be learning differentiation formulas shortly.

After taking f(x+h)-f(x)/h I came up with -6x^2-6xh-2h^2-4x^2/h-4/h, then plugging in 0 for h to take the lim h->0 I come up with -6x^2.

I have a couple more sections of this problem to work through before I can submit my answer to see if it's correct.

Thank you for your help so far!

Edit: My answer wasn't correct, so I'm going to try and work it out from the beginning again. Any suggestions? I think I see what I did (didn't work out the 4(x+h)^2 part properly), but we'll see.

Third attempt: -6x^2+8x, so -6a^2+8a

Mark44
#4
Feb8-09, 11:06 PM
Mentor
P: 21,397
(a) Find the slope of the tangent to the curve at the point where x = a.

You need parentheses.
After taking f(x+h)-f(x)/h I came up with -6x^2-6xh-2h^2-4x^2/h-4/h,
First off, its (f(x + h) - f(x))/h, not f(x+h)-f(x)/h. These are different. And with your result, I'm not sure even how to interpret it.

Let me break it down for you:
1. Calculate f(x + h)

2. Calculate f(x + h) - f(x)

3. Calculate [f(x + h) - f(x)]/h

4. Take the limit, as h approaches 0, of [f(x + h) - f(x)]/h

In your reply, please provide the results you get from each step above.
Pondera
#5
Feb8-09, 11:20 PM
P: 13
1. -2x^3+4x^2-6x^2h-6xh^2+8xh-2h^3+4h^2+4
2. -6x^2h-6xh^2+8xh-2h^3+4h^2
3. -6x^2-6xh+8x-2h^2+4h
4. -6x^2+8x
Mark44
#6
Feb8-09, 11:31 PM
Mentor
P: 21,397
Bingo and congratulations!
In the last line, what you have is f'(x).
What is f'(a)? That is the slope of the tangent line to the curve at x = a.

Some advice to make your work more readable on this forum -- use parentheses where they're needed, and space things out. You're more likely to get help if people don't have to work so hard to understand what you're trying to do.

1. -2x^3 + 4x^2 - 6hx^2 - 6xh^2 + 8xh - 2h^3 + 4h^2 + 4
2. -6hx^2 - 6xh^2 + 8xh - 2h^3 + 4h^2
3. -6x^2 -6xh + 8x - 2h^2 + 4h
4. -6x^2+8x

Notice that I moved h in the first two lines, since -6x^2h isn't as clear as -6hx^2, and makes the reader think that maybe h is in the exponent.
Pondera
#7
Feb8-09, 11:37 PM
P: 13
Isn't f'(a) just -6a^2+8a?

Thank you again for the help and the suggestions on formatting!

Edit: Apparently it is, and I was able to get the other parts of the problem (finding the equation of the tangent line given a point) easily enough. I really can't thank you enough!
HallsofIvy
#8
Feb9-09, 07:57 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
Not just formatting on this forum- your teacher will appreciate clearer writing too!

The tangent line you want has slope f'(a)= -6a^2+ 8a and passes through the point (a, f(a))= (a, 4 + 4a^2 - 2a^3). Do you know the "slope, point" formula for a line?
The Dagda
#9
Feb9-09, 09:17 AM
P: 266
Quote Quote by HallsofIvy View Post
Not just formatting on this forum- your teacher will appreciate clearer writing too!

The tangent line you want has slope f'(a)= -6a^2+ 8a and passes through the point (a, f(a))= (a, 4 + 4a^2 - 2a^3). Do you know the "slope, point" formula for a line?
Ah now I know why all my teachers hated me!

I still can't write neatly, it's something other people do. Lord praise the word processor.

That equation is the fundamental of differentiation, all of the various rules can be linked to it in one way or another. That's why you always learn it first, once you get why that works, then it makes the actual differentiation more readily and intuitively understandable.

That said god bless the rules eh. [itex]nx^{n-1}[/itex]


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