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

[frasifrasi]

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)?

 

[frasifrasi]

Also, how do I evaluate lim n --> infinity of [ln(10+x) - ln(5+x)]?

 

[rock.freak667]

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,$\frac{dy}{dx}$ and you will get the gradient of the tangent at (0,9).


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


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

and here is a hint:
$$lim_{n\rightarrow \infty} log_a [f(x)] = log_a [lim_{n\rightarrow \infty} f(x)]$$

 

[HallsofIvy]

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: $f'(x_0)$. The equation of the tangent line at $(x_0, f(x_0)$ is $y= f'(x_0)(x- x_0)$. 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 $ln(10+x)- ln(5+ x)= ln((10+x)/(5+x))$ and, since ln(x) is a continuous function, that is the ln of the limit of (10+x)/(5+x). Can you do that?

 

[frasifrasi]

Last time I checked, yes!

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

How would I find the y-intercept?

 

[frasifrasi]

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...

 

[HallsofIvy]

Last time I checked, yes!

And for the first question, I have to get the derivatie and evaluate it at $y= fsingle-quote(x_0)(x- x_0)$, 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
$$\frac{(-5)^n}{n5^n}$$?

What can you say about convergence of
$$\frac{(-1)^n}{n}$$?

That's obviously what your series reduces to.

 

[frasifrasi]

it converges conditionally.