A calc question - Tilted Forum Project Discussion Community

Stompy · 05-19-2004, 05:42 AM

Well, I guess it's more like an algebra problem since I understand the calc just fine.

I need to find an equation of the tangent line to the curve at the given point:

y = 1 + 2x - x^3 @ P(1 , 2)

To do this, I first need to find the slope with the following method: (f(x) - f(a)) / (x - a)

( (1 + 2x - x^3) - (2) ) / (x - 1)

becomes..

(2x - x^3 - 1) / (x - 1)

Top and bottom equal 0 if you plug in 1, so that means there's a way to factor them even further, correct? Meaning... (x + 1) * (?) = 2x - x^3 - 1 ?

The book says the answer is y = -x + 3, but I can't even get past simplifying (2x - x^3 - 1) / (x - 1) to get the slope.

Does x - 1 factor out of the numerator, or am I going about this all wrong?

Thanks for the help

fckm · 05-19-2004, 08:32 AM

the slope of the tangent line at the point is equal to the derivative at the point.
for:
y=1=2x-x^3,

dy/dx= 2-3x^2

evaluate at x=1
dy/dx=-1

the "slope-intercept" form of a line is:

y=mx+b

we found m=-1

and we need this line to go through the point (1,2)

plug in y=1, x=1:

2=-1+b

gets you b=3.

therefore, the tangent line has the form:

y=-1x+3

The formula that you have, (f(x) - f(a)) / (x - a), is sorta like an average slope. To get the instantaneous slope at a point, you must take the limit as x->a. Your formula then becomes the definition of a derivative (at a point x=a).

Rereading your post, you probably know this already. To take the limit, you either factor (i can't remember how to factor cubics right now, use the SOAP rule), or just use L'Hopital's rule.

EDIT: Matlab says 2x-x^3-1 = -(x-1)(x^2+x-1).
So you were on the right track. after you factor this and cancel with the bottom, plugging in x=1, you get a slope of -1.

Giltwist · 05-19-2004, 08:41 AM

Using polynomial division,

(2x - x^3 - 1) / (x - 1) = -x^2-x+1

Plugging in x=1 from the point (1,2)

(2x - x^3 - 1) / (x - 1) = -1

Therefore the slope of the tangent line is -1. Since all lines have the format y=mx+b where m is the slope, we now have

y=-x+b

using the point (1,2) again

2=-1+b

b=3

so the equation of the tangent line is

y=-x+3

Stompy · 05-19-2004, 08:42 AM

Thanks a ton!

I also figured out (eventually) the point where I was stuck:

(2x - x^3 - 1) / (x - 1)

Can't really factor from that, so I just multiplied everything by the conjucate of the denominator (x + 1) .. worked!

The concepts of calc are so easy, but it's the algebra I get stuck with... there's always an obvious way out when doing a complex problem, but it takes me FOREVER to find it..

[edit]
Nevermind, I'm all mixed up on that one. The conjugate worked only on the numerator, which gave -1, but jumped the gun without working the bottom, which still came out to 0, since I KNEW to expect -1.. so I got all "oh good, I finally found it.." when I didn't.

Working the derivative through AGAIN gave me -1, so I think I just really f'd something up in my algebra the first few times.

pig · 05-19-2004, 08:42 AM

Stompy,

Do you have to use that formula for the derivative? If not, you can just take the derivative of the original formula, and once you have the slope, you can use the equation of a line "y=mx+b" and the known point (1,2) to get the equation. If you have to use the formula, apply L'Hoptal's rule (take the derivate of the top and bottom of your expression) and plug in the point. You'll get the correct slope, and then you can find the equation of the line.

pig · 05-19-2004, 08:43 AM

crap - nevermind. I see it's been handled. That was quick.

Stompy · 05-19-2004, 10:47 AM

That's the screwed up thing about these problems... each and every problem on this page is nearly impossible to do.

It takes me like hours for each problem and I don't know wtf the issue is.

For example, the same problem above except with:

(x-1)/(x-2) at the point (3, 2).

So I try to take the derivative...

Code:

1.

(x+h) - 1      x - 1
-----------  -  ------
(x+h) - 2      x - 2

2. 

Since denoms are different, combine 'em.

(x + h - 1) (x - 2)    (x - 1) (x + h - 2)
--------------------- -  ---------------------
(x + h - 2) (x - 2)    (x + h - 2) (x - 2)


(x + h - 1) (x - 2) or
 x^2 + hx + 2 - 2x - 2h + 2 

minus 

(x - 1) (x + h - 2) or
 x^2 + hx + 2 - x - h + 2

  x^2 + hx + 2 - 2x - 2h + 2 
- (x^2 + hx + 2 - x - h + 2)

x^2 cancels out
hx cancels out
both 2s cancel out
leaving..

-2x + x
-2h + h

or.. -x - h

you get:

          -x - h                        1
---------------------------- *   ---
x^2 + hx - 2x - 2h + 8        h


3. h cancels out leaving...

         -x - 1
-------------------------
x^2 + 0 - 2x - 0 + 8

or...

     -x - 1
----------------
x^2 - 2x + 8

using 3, we get..

     -(3) - 1
--------------------
(3)^2 - 2(3) + 8

-4
---
11

y - 2 = - 4/11(x - 3)
y = -4/11x - (12/11) + 22/11
y = -4/11x + 10/11

The book has goddamn y = -x + 5. WTF?

This stuff pisses me off. I spent hours on that first one, hours on this.. and I'm 2 problems through this 30 + problem homework paper due tonight.

KnifeMissile · 05-19-2004, 01:02 PM

You simply made an algebraic error in determining the derivative. Do it again, from scratch, without looking at what you've already done and, hopefully, you will find a different and correct answer...

pig · 05-19-2004, 01:03 PM

Stompy

Use this : Derivative of Hi/Ho = (HodHi - HidHo)/Ho^2.

take the derivative using that and you'll get the answer.

pig · 05-19-2004, 01:04 PM

Or what Knife Missile said.

fckm · 05-19-2004, 08:01 PM

i take it you guys haven't gone over chain rule and all that nice stuff. Once you stop using the limit formula to calculate derivatives, things become much easier.
remember some simple facts, and it'll help you a lot.

d/dx ( x^a) = a*x^(a-1)

d/dx [f(x)*g(x)] = [d/dx f(x) ]g(x) + f(x)[d/dx g(x)]

d/dx[f(x) + g(x)] = d/dx f(x) + d/dx g(x)

d/dx [f(g(x))] = f'(g(x)) * g'(x)

where f(x) and g(x) are functions of x and ' signifies a derivative.

now you can take any derivative of any combination of polynomial functions.
so,

y(x)=(x-1)/(x-2)
=(x-1)*(x-2)^(-1)

dy/dx=1*(x-2)^(-1) + (x-1)*(-1)(x-2)^(-2)*1

dy/dx @x=3 -> (3-2)^(-1)+(3-1)*(-1)(3-2)^(-2)
=1+(2)*(-1) = -1

Stompy · 05-20-2004, 04:01 AM

Thanks for all the help!

I'm not quite sure what I was thinkin yesterday, but I tried the probs again after I had a few more hrs sleep and I didn't seem to have any issue with them, which is VERY odd. It's as if I had someone else's brain in my head..

Just one of those days, I guess

Thanks again

05-19-2004, 05:42 AM	#1 (permalink)
Stompy Banned from being Banned Location: Donkey	A calc question Well, I guess it's more like an algebra problem since I understand the calc just fine. I need to find an equation of the tangent line to the curve at the given point: y = 1 + 2x - x^3 @ P(1 , 2) To do this, I first need to find the slope with the following method: (f(x) - f(a)) / (x - a) ( (1 + 2x - x^3) - (2) ) / (x - 1) becomes.. (2x - x^3 - 1) / (x - 1) Top and bottom equal 0 if you plug in 1, so that means there's a way to factor them even further, correct? Meaning... (x + 1) * (?) = 2x - x^3 - 1 ? The book says the answer is y = -x + 3, but I can't even get past simplifying (2x - x^3 - 1) / (x - 1) to get the slope. Does x - 1 factor out of the numerator, or am I going about this all wrong? Thanks for the help __________________ I love lamp.

05-19-2004, 08:32 AM	#2 (permalink)
fckm Insane Location: Ithaca, New York	the slope of the tangent line at the point is equal to the derivative at the point. for: y=1=2x-x^3, dy/dx= 2-3x^2 evaluate at x=1 dy/dx=-1 the "slope-intercept" form of a line is: y=mx+b we found m=-1 and we need this line to go through the point (1,2) plug in y=1, x=1: 2=-1+b gets you b=3. therefore, the tangent line has the form: y=-1x+3 The formula that you have, (f(x) - f(a)) / (x - a), is sorta like an average slope. To get the instantaneous slope at a point, you must take the limit as x->a. Your formula then becomes the definition of a derivative (at a point x=a). Rereading your post, you probably know this already. To take the limit, you either factor (i can't remember how to factor cubics right now, use the SOAP rule), or just use L'Hopital's rule. EDIT: Matlab says 2x-x^3-1 = -(x-1)(x^2+x-1). So you were on the right track. after you factor this and cancel with the bottom, plugging in x=1, you get a slope of -1. __________________ And if you say to me tomorrow, oh what fun it all would be. Then what's to stop us, pretty baby. But What Is And What Should Never Be. Last edited by fckm; 05-19-2004 at 08:36 AM..

05-19-2004, 08:42 AM	#4 (permalink)
Stompy Banned from being Banned Location: Donkey	Thanks a ton! I also figured out (eventually) the point where I was stuck: (2x - x^3 - 1) / (x - 1) Can't really factor from that, so I just multiplied everything by the conjucate of the denominator (x + 1) .. worked! The concepts of calc are so easy, but it's the algebra I get stuck with... there's always an obvious way out when doing a complex problem, but it takes me FOREVER to find it.. [edit] Nevermind, I'm all mixed up on that one. The conjugate worked only on the numerator, which gave -1, but jumped the gun without working the bottom, which still came out to 0, since I KNEW to expect -1.. so I got all "oh good, I finally found it.." when I didn't. Working the derivative through AGAIN gave me -1, so I think I just really f'd something up in my algebra the first few times. __________________ I love lamp. Last edited by Stompy; 05-19-2004 at 08:59 AM..

05-19-2004, 08:42 AM	#5 (permalink)
pig pigglet pigglet Location: Locash	Stompy, Do you have to use that formula for the derivative? If not, you can just take the derivative of the original formula, and once you have the slope, you can use the equation of a line "y=mx+b" and the known point (1,2) to get the equation. If you have to use the formula, apply L'Hoptal's rule (take the derivate of the top and bottom of your expression) and plug in the point. You'll get the correct slope, and then you can find the equation of the line. __________________ You don't love me, you just love my piggy style

05-19-2004, 08:43 AM	#6 (permalink)
pig pigglet pigglet Location: Locash	crap - nevermind. I see it's been handled. That was quick. __________________ You don't love me, you just love my piggy style

05-19-2004, 08:41 AM	#3 (permalink)
Giltwist Insane Location: Pennsylvania	Using polynomial division, (2x - x^3 - 1) / (x - 1) = -x^2-x+1 Plugging in x=1 from the point (1,2) (2x - x^3 - 1) / (x - 1) = -1 Therefore the slope of the tangent line is -1. Since all lines have the format y=mx+b where m is the slope, we now have y=-x+b using the point (1,2) again 2=-1+b b=3 so the equation of the tangent line is y=-x+3

05-19-2004, 01:02 PM	#8 (permalink)
KnifeMissile Location: Waterloo, Ontario	You simply made an algebraic error in determining the derivative. Do it again, from scratch, without looking at what you've already done and, hopefully, you will find a different and correct answer...

05-19-2004, 01:03 PM	#9 (permalink)
pig pigglet pigglet Location: Locash	Stompy Use this : Derivative of Hi/Ho = (HodHi - HidHo)/Ho^2. take the derivative using that and you'll get the answer. __________________ You don't love me, you just love my piggy style

05-19-2004, 01:04 PM	#10 (permalink)
pig pigglet pigglet Location: Locash	Or what Knife Missile said. __________________ You don't love me, you just love my piggy style

05-19-2004, 08:01 PM	#11 (permalink)
fckm Insane Location: Ithaca, New York	i take it you guys haven't gone over chain rule and all that nice stuff. Once you stop using the limit formula to calculate derivatives, things become much easier. remember some simple facts, and it'll help you a lot. d/dx ( x^a) = ax^(a-1) d/dx [f(x)g(x)] = [d/dx f(x) ]g(x) + f(x)[d/dx g(x)] d/dx[f(x) + g(x)] = d/dx f(x) + d/dx g(x) d/dx [f(g(x))] = f'(g(x)) * g'(x) where f(x) and g(x) are functions of x and ' signifies a derivative. now you can take any derivative of any combination of polynomial functions. so, y(x)=(x-1)/(x-2) =(x-1)(x-2)^(-1) dy/dx=1(x-2)^(-1) + (x-1)(-1)(x-2)^(-2)1 dy/dx @x=3 -> (3-2)^(-1)+(3-1)(-1)(3-2)^(-2) =1+(2)(-1) = -1 __________________ And if you say to me tomorrow, oh what fun it all would be. Then what's to stop us, pretty baby. But What Is And What Should Never Be.

05-20-2004, 04:01 AM	#12 (permalink)
Stompy Banned from being Banned Location: Donkey	Thanks for all the help! I'm not quite sure what I was thinkin yesterday, but I tried the probs again after I had a few more hrs sleep and I didn't seem to have any issue with them, which is VERY odd. It's as if I had someone else's brain in my head.. Just one of those days, I guess Thanks again __________________ I love lamp.