Intersection of tangent line and curve

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BeeCuz

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I have a formula y=log(x)/log(0.9), and the graph of it looks like this:

Untitled.png

I wish to find the intersection of this curve and a tangent line that appears like so in this rough sketch

Untitled2.png
But of course the scales of the axes are quite different, so it's not actually a line with a slope of -1, it just sort of looks that way in this image.

How can I:
1. Figure out what that slope actually is
2. Figure out the coordinates of the intersection?
 
How do your scales compare? It looks like 5 up is the same length as 0.1 across. So the ratio of x-scale to y-scale is 0.1 : 5 = 1:50. Is that correct?
 
BeeCuz said:
I have a formula y=log(x)/log(0.9), and the graph of it looks like this:

View attachment 30916

I wish to find the intersection of this curve and a tangent line that appears like so in this rough sketch

View attachment 30917
But of course the scales of the axes are quite different, so it's not actually a line with a slope of -1, it just sort of looks that way in this image.

How can I:
1. Figure out what that slope actually is
2. Figure out the coordinates of the intersection?
Click to expand...
What the slope is depends on where it is tangent. And the intersection is the point of tangency. If you don't know that point, then what do you know about the tangent line? There must be some known information!

Please state the actual, entire problem you are working on.

Did you differentiate the function? If you know the point is at x=0.2, did you plug that into both the function and its derivative?
 
The slope is given. It's -50. The question is what are the coordinates.

I can manufacture justifications for my own ignorance all by myself all day long, don't need help with that thanks.
 
BeeCuz said:
The slope is given. It's -50. The question is what are the coordinates.

I can manufacture justifications for my own ignorance all by myself all day long, don't need help with that thanks.
Click to expand...
Do you remember that the slope is equal to the derivative of the function?
 
BeeCuz said:
The slope is given. It's -50. The question is what are the coordinates.
Click to expand...
So, the problem you are trying to solve is, I think,

Given the curve y=log(x)/log(0.9), what is the point where the tangent line has slope -50?​

This wasn't clear from the original question (especially, what facts we are to assume from the picture, as it's easier to guess the coordinates of the point than the slope).

Since this site is for helping people solve problems, not just giving answers, we need to know what you do know, so we can help you use it. You mentioned dy/dx, so you clearly do know that the slope is the derivative; shall we assume you need help finding the derivative of your function?
 
blamocur said:
Do you remember that the slope is equal to the derivative of the function?
Click to expand...
Thank you for the information.

Yes, I've seen someone else say "find the derivative" as well. So apparently it's possible to solve this problem with calculus. Which is why I came to this forum on calculus, not because I know how to do calculus but because the problem appears to be one that calculus can solve.
 
BeeCuz said:
Thank you for the information.

Yes, I've seen someone else say "find the derivative" as well. So apparently it's possible to solve this problem with calculus. Which is why I came to this forum on calculus, not because I know how to do calculus but because the problem appears to be one that calculus can solve.
Click to expand...
What was your method besides calculus?
 
BeeCuz said:
That's for other people to consider, in other forums.

Given y=log(x)/log(9/10) and a tangent line with a slope of -50, what are the coordinates of their intersection?
Click to expand...
So now that you know the method to solve the problem using calculus (response #8) - solve it and share your work.
 
Subhotosh Khan said:
So now that you know the method to solve the problem using calculus
Click to expand...
Actually that's a shorthand phrase someone spoke, with unwritten details; a link to additional necessary information but without the link.

But since you're a staff member maybe you know how to translate that phrase into a sequence of mathematical actions.
Or if unwilling, then maybe someone can point to a video that does show that process in action.
 
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BeeCuz said:
Actually that's a shorthand phrase someone spoke, with unwritten details; a link to additional necessary information but without the link.

But since you're a staff member maybe you know how to translate that phrase into a sequence of mathematical actions.
Or if unwilling, then maybe someone can point to a video that does show that process in action.
Click to expand...
In response #8, there is no link - there is a method of solution suggested.

The method suggested in there should be clear to a student of Calculus. If it is not clear to you, may be you should learn/brush-up the relationship between the gradient/slope of a function, at a given point, and the derivative of the function. Do a Google search and let us know what you find.
 
The important information in the plot is the y-intercept. I think the interection to the curve is purposefully left unclear because that is information you do not need. There can only be one line that passes through (0,25) and is also tangent to the curve.
if the point of interection is [imath](x_0,y_0)[/imath] then the slope of the line is [imath] m = \frac{y_0-25}{x_0}[/imath]. You can now get another expression for the slope at [imath](x_0,y_0)[/imath] by differentiating the curve equation, set then equal.

[math] m = \frac{y_0-25}{x_0} = \frac{dy}{dx}|_{x=x_0} .......... (1)[/math]
also note that since the intersection point lies on the curve,
[math]y_0 = \frac{log(x_0)}{log(0.9)} .............(2)[/math]From the above two equations, you can calculate the intersection point.

Once you have the intersection point, you can find the slope quite easily.
 
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BeeCuz said:
maybe someone can point to a video that does show that process
Click to expand...
Hi BeeCuz. The process of differentiating log(x) depends upon the base. Many people express the base-10 logarithm using the name 'log', but some people use that name for base-e logarithms, instead. Before I look for an introductory video covering the process, I'd like to know whether the 'log' in your exercise denotes a base-10 or base-e logarithm.

Either base may be used, but the choice of base leads to different steps. Differentiating [imath]\text{log}_{10}(x)[/imath] takes place via the change-of-base formula, by first switching to [imath]\text{log}_{e}(x)[/imath]. I would search for a video to match.

Do you know the base, in the function you were given?

:)

PS: I always write log(x) for base-10 and ln(x) for base-e.

[imath]\;[/imath]
 
Fermat2718 said:
The important information in the plot is the y-intercept. I think the interection to the curve is purposefully left unclear because that is information you do not need. There can only be one line that passes through (0,25) and is also tangent to the curve.
if the point of interection is [imath](x_0,y_0)[/imath] then the slope of the line is [imath] m = \frac{y_0-25}{x_0}[/imath]. You can now get another expression for the slope at [imath](x_0,y_0)[/imath] by differentiating the curve equation, set then equal.

[math] m = \frac{y_0-25}{x_0} = \frac{dy}{dx}|_{x=x_0} .......... (1)[/math]
also note that since the intersection point lies on the curve,
[math]y_0 = \frac{log(x_0)}{log(0.9)} .............(2)[/math]From the above two equations, you can calculate the intersection point.

Once you have the intersection point, you can find the slope quite easily.
Click to expand...
One of my frustrations with this thread was that the actual problem was never clearly stated (in addition to the fact that the OP never quite said what knowledge he has on which to base a solution, or what part of the work he found difficult).

I hadn't considered this as an interpretation of the problem: Find the point on the curve whose tangent line has y-intercept 25. I like that, because 25 is the only number that can be read clearly from the graph. I just wish the question had been worded in a way that made that clear.

Assuming that is correct, it's a nice problem, and we do find that the solution agrees with the graph, while being more precise.

Interestingly, the graph (and the solution) is the same whether "log" in the problem is the common or natural log, as either way, the function is just [imath]\log_{0.9}x[/imath]. I had assumed it was the common log because the graph agreed with what I got when I took it that way; I missed the fact that it's the same either way. In solving it just now, I took it as the natural log, only because that's easier.

And if the OP's difficulty was in not knowing how to differentiate the common log, this would help.
 
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