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Help - tangent to y = 9e^(2x)cos(pi*x)

  • Thread starter frasifrasi
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  • #1
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Q: find the tangent to y = 9e^(2x)cos(pi*x) at (0,9)

I know how to do the tangent for a polar coordinate at an angle, but how find it at (0,9)?
 

Answers and Replies

  • #2
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Also, how do I evaluate lim n --> infinity of [ln(10+x) - ln(5+x)]?
 
  • #3
rock.freak667
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For the first question...
to find the equation of a tangent, you need the point and the gradient of the tangent at that point. So you already have the point (0,9).

Sub x=0 into the gradient function,[itex]\frac{dy}{dx}[/itex] and you will get the gradient of the tangent at (0,9).


For your second problem:
[tex]lim_{n\rightarrow \infty} [ln(10+x) - ln(5+x)][/tex]


is there any way you can simplify ln(10+x) - ln(5+x)?

and here is a hint:
[tex]lim_{n\rightarrow \infty} log_a [f(x)] = log_a [lim_{n\rightarrow \infty} f(x)][/tex]
 
  • #4
HallsofIvy
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Q: find the tangent to y = 9e^(2x)cos(pi*x) at (0,9)

I know how to do the tangent for a polar coordinate at an angle, but how find it at (0,9)?
I'm not even sure what a "tangent for a polar coordinate at an angle" is, but finding the tangent to a curve at a given point is one of the first things I learned in calculus!

Or are you talking about the tangent of an angle? Surely not!

To find the tangent line to a curve at a given point, first find the derivative there: [itex]f'(x_0)[/itex]. The equation of the tangent line at [itex](x_0, f(x_0)[/itex] is [itex]y= f'(x_0)(x- x_0)[/itex]. The only "hard part" here is finding the derivative- and it's not all that hard. Use the product law along with the chain rule.

Also, how do I evaluate lim n --> infinity of [ln(10+x) - ln(5+x)]?
Well, you certainly know, I hope, that [itex]ln(10+x)- ln(5+ x)= ln((10+x)/(5+x))[/itex] and, since ln(x) is a continuous function, that is the ln of the limit of (10+x)/(5+x). Can you do that?
 
  • #5
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Last time I checked, yes!

And for the first question, I have to get the derivatie and evaluate it at [itex]y= fsingle-quote(x_0)(x- x_0)[/itex], correct?

How would I find the y-intercept?
 
Last edited:
  • #6
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One more thing, how do i show that (-5)^n over n*5^n

converges? By the alternating series test? I am not sure how to manupulate the (-5)^n on the top.


Sorry about the stupid questions, I am trying to prepare for the test.

EDIT: i guess It's just (-1)^n(5)^n...
 
Last edited:
  • #7
HallsofIvy
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Last time I checked, yes!

And for the first question, I have to get the derivatie and evaluate it at [itex]y= fsingle-quote(x_0)(x- x_0)[/itex], correct?

How would I find the y-intercept?
No, you have to evaluate the derivative at x= 0. I can see no reason why you would want to find the y-intercept.

One more thing, how do i show that (-5)^n over n*5^n

converges? By the alternating series test? I am not sure how to manupulate the (-5)^n on the top.


Sorry about the stupid questions, I am trying to prepare for the test.

EDIT: i guess It's just (-1)^n(5)^n...
You mean
[tex]\frac{(-5)^n}{n5^n}[/tex]?

What can you say about convergence of
[tex]\frac{(-1)^n}{n}[/tex]?

That's obviously what your series reduces to.
 
  • #8
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it converges conditionally.
 

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