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Determine the equation for the the tangent

  • Thread starter Gliese123
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  • #1
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Homework Statement


Determine the equation for the the tangent-to the curve:
y=3 sin 2x - cos 2x
if x=3∏/4


Homework Equations


So I thought I might get the y?

y=3 sin 2(3∏/4) - cos 2(3∏/4)
y= ~0.25 - ~1 ≈ 0.75

k(?) =
m(?) =
Then what?
Please help :(

The Attempt at a Solution



The answer is y = -2x + 3∏/2 - 3
 
Last edited:

Answers and Replies

  • #2
rock.freak667
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For y=3 sin2x - cos2x, how would you find the gradient function i.e. the function which gives the gradient at any point x?
 
  • #3
eumyang
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Homework Statement


Determine the equation for the the tangent-to the curve:
y=3 sin 2x - cos 2x
if x=3π/4


Homework Equations


So I thought I might get the y?

y=3 sin 2(3π/4) - cos 2(3π/4)
y= ~0.25 - ~1 ≈ 0.75
This isn't correct. If you are using a calculator, are you sure you are in radian mode?

Once you get the correct y coordinate in the point, find the derivative of y. Plug x = 3π/4 into the derivative to get the slope of the tangent line. Finally, plug in the point and slope into the point-slope form of the equation of a line.
 
  • #4
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For y=3 sin2x - cos2x, how would you find the gradient function i.e. the function which gives the gradient at any point x?
I don't know really. Should I deriving?
 
  • #5
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This isn't correct. If you are using a calculator, are you sure you are in radian mode?

Once you get the correct y coordinate in the point, find the derivative of y. Plug x = 3π/4 into the derivative to get the slope of the tangent line. Finally, plug in the point and slope into the point-slope form of the equation of a line.
No I think its adjustments are set to angles. Is it possible to do it without setting it to radians?
 
  • #6
Bacle2
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I don't know really. Should I deriving?
Do you know the definition of the derivative in terms of the tangent line?

This is what I think roc.freak was getting at.
 
  • #7
eumyang
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No I think its adjustments are set to angles. Is it possible to do it without setting it to radians?
You could also memorize the sine and cosine of special angles.
[itex]y = 3 \sin \left(2 \cdot \frac{3\pi}{4}\right) - \cos \left(2 \cdot \frac{3\pi}{4}\right)[/itex]
[itex]y = 3 \sin \left(\frac{3\pi}{2}\right) - \cos \left(\frac{3\pi}{2}\right)[/itex]
3π/2 is one such special angle. What is
[itex]\sin \left(\frac{3\pi}{2}\right)[/itex]
and
[itex]\cos \left(\frac{3\pi}{2}\right)[/itex]?
 
  • #8
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You could also memorize the sine and cosine of special angles.
[itex]y = 3 \sin \left(2 \cdot \frac{3\pi}{4}\right) - \cos \left(2 \cdot \frac{3\pi}{4}\right)[/itex]
[itex]y = 3 \sin \left(\frac{3\pi}{2}\right) - \cos \left(\frac{3\pi}{2}\right)[/itex]
3π/2 is one such special angle. What is
[itex]\sin \left(\frac{3\pi}{2}\right)[/itex]
and
[itex]\cos \left(\frac{3\pi}{2}\right)[/itex]?
When I put that on the calculator I get:
sin to be -0.99... and cos to be -2.38 × 10-3 :cry:
And it's set to radians.
 
  • #9
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When I put that on the calculator I get:
sin to be -0.99... and cos to be -2.38 × 10-3 :cry:
And it's set to radians.
Should I take 3 × -0.9999 = -3. Is that the gradient?
 
  • #10
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Do you know the definition of the derivative in terms of the tangent line?

This is what I think roc.freak was getting at.
No. Sorry. I know how to derivate sin, cos etc. though :redface:
 
  • #11
eumyang
Homework Helper
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When I put that on the calculator I get:
sin to be -0.99... and cos to be -2.38 × 10-3 :cry:
And it's set to radians.
Something is wrong with your calculator. Because
sin (3π/2) = -1 and cos (3π/2) = 0.

Should I take 3 × -0.9999 = -3. Is that the gradient?
No, it's not the gradient! It's the y-coordinate of the point where the tangent line intersects the curve. Please reread my post.
 
  • #12
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Something is wrong with your calculator. Because
sin (3π/2) = -1 and cos (3π/2) = 0.


No, it's not the gradient! It's the y-coordinate of the point where the tangent line intersects the curve. Please reread my post.

Oh sorry!:redface: Hm. Okay so is y= -1 then? Sigh.. I don't know?
Should I derivate the function and put -1 = (the derivation)? (And include x)?
 
  • #13
eumyang
Homework Helper
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Oh sorry! Hm. Okay so is y= -1 then? Sigh.. I don't know?
No, y does not equal -1. There was a 3 in front of the sine.

Should I derivate the function and put -1 = (the derivation)? (And include x)?
No again. You should take the derivative of
[itex]y = 3 \sin 2x - \cos 2x[/itex]
[itex]y' = ...[/itex]
and then plug in x = 3π/4. y' is the slope of the tangent line at x = 3π/4.
 
  • #14
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No, y does not equal -1. There was a 3 in front of the sine.


No again. You should take the derivative of
[itex]y = 3 \sin 2x - \cos 2x[/itex]
[itex]y' = ...[/itex]
and then plug in x = 3π/4. y' is the slope of the tangent line at x = 3π/4.
So
y' = 3cos2x + 2sin2x right?
y' = 3 cos 2(3∏/4) + 2 sin 2(3∏/4) = - 4 ?
Could you lead me through please? I can't do this on my own. :confused:
 
  • #15
eumyang
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So
y' = 3cos2x + 2sin2x right?
No. You are forgetting the chain rule.
 
  • #16
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No. You are forgetting the chain rule.
I give up... Thanks very much for the help though.. :)
 
  • #17
HallsofIvy
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The problem appears to be that you are trying to do a Calculus problem involving trig functions but have never learned Calculus or Trigonometry.
 
  • #18
Simon Bridge
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It's in "homework" ... so it looks like OP is doing a course that requires calculus and trigonometry without having learned the prerequisites.

Actually, the question would be a good entry test for any situation where a basic understanding of calc and trig are required but not taught. Like a job interview.
 
  • #19
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It's in "homework" ... so it looks like OP is doing a course that requires calculus and trigonometry without having learned the prerequisites.

Actually, the question would be a good entry test for any situation where a basic understanding of calc and trig are required but not taught. Like a job interview.
Yes. It's in "Homework" which means that I'm no pro on this one. I just started with the concept of derivations in trigonometry. I get that there're a lot of rules when one derivate functions including trigonometry. What have caused me to fail the calulations is because my calculator seem to not be set to the primary settings. I hoped that someone could more or less tell me the finnish. That would be helpful, not just telling me what's wrong all the time. :smile:
 
  • #20
Simon Bridge
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Yes. It's in "Homework" which means that I'm no pro on this one. I just started with the concept of derivations in trigonometry.
You mean, "derivatives of trigonometric functions"?
Or, that your course is on the concepts of differentiating trigonometric functions?
I get that there're a lot of rules when one derivate functions
"when one differentiates functions"
including trigonometry. What have caused me to fail the calulations is because my calculator seem to not be set to the primary settings.
You don't need a calculator at all so this is not true.
I hoped that someone could more or less tell me the finnish. That would be helpful, not just telling me what's wrong all the time.
That would be against the forum rules.

The idea is that you do your homework - we can help out where you get stuck but we will not do the work for you. Thing is, you have been stuck on everything. You do not know how a sine function varies with angle, and you don't know the chain rule. Both of these things were in an earlier part of your coursework.

Therefor: you have to go back over your course notes.
 
  • #21
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That would be against the forum rules.

The idea is that you do your homework - we can help out where you get stuck but we will not do the work for you. Thing is, you have been stuck on everything. You do not know how a sine function varies with angle, and you don't know the chain rule. Both of these things were in an earlier part of your coursework.

Therefor: you have to go back over your course notes.
Well, don't get me wrong now, but why did I even bother to write the question here? Math isn't easy to understand, that's for sure but wow, this takes the prize of impossibility. I respect the forum rules but I guess that I need to be 99% knowing of the statement's problem to even write it here. I'll have that in mind. Thank you..
 
  • #22
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Okay dudes I got the answer:
y= 3sin 2x - cos 2x (x=3∏/4) =>
y= 3 sin 2(3∏/4) - cos 2(3∏/4)
y=-3

y' = 6cos2x - 2sin 2x
y' = 6cos2(3∏/4) - 2sin 2(3∏/4)
y'= -2 (-2= gradient I guess??)

y= kx + m

-3= -2(3∏/4) + m
m = -2(3∏/4) -3 m = 3∏/2 - 3 short
So:

y = -2x + (3∏/2) - 3

It looks right at least?
 
Last edited:
  • #23
Simon Bridge
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y' = 6cos2x - 2sin 2x
derivative of cosine is -sine

you can figure it out from a sketch of the cosine curve and looking at the gradient at different places. The cosine starts out at y=1 with gradient of 0 ... as x increases, the gradient becomes negative. When the cosine first crosses the axis, the gradient is -1/2. And so on. You don't need to memorize this if you know what a gradient is and you can sketch, even very roughly, the cosine function.

If you don't take the trouble to understand things you will keep repeating the same mistakes.

why did I even bother to write the question here?
Turn that around - why do we even bother trying to help you?

When someone does not want to learn, does not come with even the basic concepts around the subject so we basically have to teach them maths - for free? This is a two-way street. You've posted enough here to know the score by now.
Math isn't easy to understand, that's for sure but wow, this takes the prize of impossibility. I respect the forum rules but I guess that I need to be 99% knowing of the statement's problem to even write it here.
Actually, if the place you get stuck is understanding the problem, then we can and will unstick you. That is not the same a doing the problem for you.

The principle is that you learn best by doing the work yourself. I know it is hard. We all do. We all did it the same way, tried shortcuts and walkthrough's, learned the hard way that this is a mistake, and now pass this on for the next generation to ignore ;)

There are services which provide the kind of assistance you asked for ... but you usually have to pay for them.

One of the neat things about offering free help is that you get to offer the help that people need rather than the help that they ask for. When you pay for my help, then you get to tell me what to do.

You don't have to take the advise offered here. You don't even have to like it.
 
  • #24
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Actually, if the place you get stuck is understanding the problem, then we can and will unstick you. That is not the same a doing the problem for you.

The principle is that you learn best by doing the work yourself. I know it is hard. We all do. We all did it the same way, tried shortcuts and walkthrough's, learned the hard way that this is a mistake, and now pass this on for the next generation to ignore ;)

There are services which provide the kind of assistance you asked for ... but you usually have to pay for them.

One of the neat things about offering free help is that you get to offer the help that people need rather than the help that they ask for. When you pay for my help, then you get to tell me what to do.

You don't have to take the advise offered here. You don't even have to like it.
Yeah. Well. I thank you for your effort :) Don't misstaken me :) What I got seemed to be right. That what's matters :)
 
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