# wow linear approximation

• Nov 3rd 2007, 08:24 AM
aslan
wow linear approximation
- its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:

Use differentials (linear approximation) to estimate https://webwork.rutgers.edu/webwork2...1296499281.png. Base the approximation at https://webwork.rutgers.edu/webwork2...f58ee651a1.png.

ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?
• Nov 3rd 2007, 08:41 AM
topsquark
Quote:

Originally Posted by aslan
- its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:

Use differentials (linear approximation) to estimate https://webwork.rutgers.edu/webwork2...1296499281.png. Base the approximation at https://webwork.rutgers.edu/webwork2...f58ee651a1.png.

ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?

First, $\frac{d}{dx}sin(x) = cos(x)$, so
$sin(x) \approx sin(x_0) + cos(x_0) \cdot (x - x_0)$
to first order, where $x_0$ is the value of the argument that you are using as an estimate.

So
$sin(-2) \approx sin \left ( -\frac{2\pi}{3} \right ) + cos \left ( - \frac{2\pi}{3} \right ) \cdot \left ( -\frac{2\pi}{3} - - 2 \right )$

$sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - 2 -- \frac{2\pi}{3} \right )$

In terms of decimals:
$sin(-2) \approx -0.913223$

The actual value of $sin(-2) = -0.909297$ so we are talking about a difference of 0.004317%.

-Dan
• Nov 3rd 2007, 08:54 AM
aslan
Quote:

Originally Posted by topsquark
First, $\frac{d}{dx}sin(x) = cos(x)$, so
$sin(x) \approx sin(x_0) + cos(x_0) \cdot (x - x_0)$
to first order, where $x_0$ is the value of the argument that you are using as an estimate.

So
$sin(-2) \approx sin \left ( -\frac{2\pi}{3} \right ) + cos \left ( - \frac{2\pi}{3} \right ) \cdot \left ( -\frac{2\pi}{3} - - 2 \right )$

$sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )$

In terms of decimals:
$sin(-2) \approx -0.818828$

The actual value of $sin(-2) = -0.909297$ so we are talking about a difference of 0.99%.

-Dan

im starting to understand this. but here:
$sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )$
wouldnt the ride side be (-2 + 2pi/3) instead of (-2pi/3 + 2) since it is (x-a)?
• Nov 3rd 2007, 08:59 AM
topsquark
Quote:

Originally Posted by aslan
im starting to understand this. but here:
$sin(-2) \approx -\frac{\sqrt{3}}{2} - \frac{1}{2} \cdot \left ( - \frac{2\pi}{3} + 2 \right )$
wouldnt the ride side be (-2 + 2pi/3) instead of (-2pi/3 + 2) since it is (x-a)?

You are correct, I switched the order, which improves the error tremendously. (I thought it looked a little high.) I have fixed this in my original post.

See, you are understanding this! (Handshake)

-Dan
• Nov 3rd 2007, 09:09 AM
aslan
Quote:

Originally Posted by topsquark
You are correct, I switched the order, which improves the error tremendously. (I thought it looked a little high.) I have fixed this in my original post.

See, you are understanding this! (Handshake)

-Dan

Thanks a lot man, yeah i think somehow linear approximation just became a lot more understandable.
• Nov 3rd 2007, 12:08 PM
CaptainBlack
Quote:

Originally Posted by aslan
- its killing me. ive been fine in calculus but for some reason linear approximation is just not clicking for me. here's the problem:

Use differentials (linear approximation) to estimate (Doh). Base the approximation at (Doh).

ive tried solving this many different ways, always different answers, always incorrect. if anyone can help, with a step by step procedure so when i see another similar problem ill be able to solve it?

Please use a different image hosting service, the one you are using
does not have acceptable security certificates for many of our computers.

RonL
• Nov 3rd 2007, 05:54 PM
aslan
oh, sorry about the imagine hosting then.

any help with this problem on a different subject:

The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

Now estimate the relative error in the calculated surface area. ________

im supposing i'd use the formula:
E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.
• Nov 3rd 2007, 08:28 PM
aslan
Quote:

Originally Posted by aslan
oh, sorry about the imagine hosting then.

any help with this problem on a different subject:

The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

Now estimate the relative error in the calculated surface area. ________

im supposing i'd use the formula:
E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.

bump
• Nov 4th 2007, 09:56 AM
aslan
Quote:

Originally Posted by aslan
oh, sorry about the imagine hosting then.

any help with this problem on a different subject:

The circumference of a sphere was measured to be 73 cm with a possible error of 0.5 cm. Use differentials to estimate the maximum error in the calculated surface area. ________

Now estimate the relative error in the calculated surface area. ________

im supposing i'd use the formula:
E(h) = f(a+h) - [f(a) + h*derivative of f(a)]
where h is the difference between x and a. but using this formula doesnt tell me anything about the maximum or relative error..

btw instead of solving this if anyone would think it was easier to just direct me to a page that explains all applications of linear approximation, id be glad. i dont have a textbook for my calc class because of money problems and for some reason its been hard for me to find sample problems like these online.

BUMP for this

equations:
C=2r*pi
S=4pi*r^2

my process was using the error for circumference to solve for "dr", and then using circumference to solve for the "r." then with those values, differentiate the surface area equation and solve for maximum error. but i am continuously getting the wrong answer.

anyone with any help?
• Nov 4th 2007, 08:02 PM
topsquark
Quote:

Originally Posted by aslan
BUMP for this

equations:
C=2r*pi
S=4pi*r^2

my process was using the error for circumference to solve for "dr", and then using circumference to solve for the "r." then with those values, differentiate the surface area equation and solve for maximum error. but i am continuously getting the wrong answer.

anyone with any help?

I'm not answering this one because I don't know what to do with the errors.

But I wanted to say this:
"Bumping" is rude! If someone is going to answer your question they will answer it their own time. Bumping to get more attention paid to your question is irritating (the whole point of it, really) and makes no sense at all. I mean, if someone was going to answer your question, they'd likely do so the first time they read it.

-Dan
• Nov 4th 2007, 08:32 PM
aslan
Quote:

Originally Posted by topsquark
I'm not answering this one because I don't know what to do with the errors.

But I wanted to say this:
"Bumping" is rude! If someone is going to answer your question they will answer it their own time. Bumping to get more attention paid to your question is irritating (the whole point of it, really) and makes no sense at all. I mean, if someone was going to answer your question, they'd likely do so the first time they read it.

-Dan

what? the whole point of bumping is to call attention to a question. this is the urgent section, i see, and if there hasnt been a response in say, 12 hours, wouldnt a person think the OP no longer needed the help? bumping lets people know the subject at hand is still a concern.

im not new to forums lol, though i am new to seeing moderators refer to bumping as rude. ridiculous.
• Nov 4th 2007, 08:51 PM
topsquark
Quote:

Originally Posted by aslan
what? the whole point of bumping is to call attention to a question. this is the urgent section, i see, and if there hasnt been a response in say, 12 hours, wouldnt a person think the OP no longer needed the help? bumping lets people know the subject at hand is still a concern.

im not new to forums lol, though i am new to seeing moderators refer to bumping as rude. ridiculous.

I'm not a mod, I simply help here.

It could simply be that no one knows how to deal with your question. In that case, why bump?

It could be that a person that can answer your question hasn't logged on yet. In that case they will see your question when they do log on. So again, why bump? This person isn't going to log in any sooner because you bumped!

As much as I'd like to not say this, there are many questions that posters have asked that never get a response. In most cases it's because no one knows how to answer the question. (In some cases, it's because the poster is such a jerk that no one wants to, but that isn't the case with you, I assure you!)

Bumping is a means to an end: the idea is to call attention to your message, therefore prioritizing your message over everyone else's. I find this kind of behavior rude. If you don't agree, so be it.

-Dan