Functions sturglle

Question

Let D(x) be the function that represents the distance from (1,1) to the point (x, f (x)) where f(x)= lnx. Find the rule for D(x) and the domain of the function D. Include a diagram (follow the guidelines for diagrams and graphs) that shows the graph of f, the point (1,1) and a line from that point to an arbitrary point (x, f (x)) on the graph of f. Explain how you determined the rule for D. 

Answer 1

Hi Monas,

 

Let's begin by answering the last question first.

 

We determine the rule of D(x) by realizing it's the equation of the formula used to determine the distance between two points:

    distance formula:  √((x₂-x₁)² + (y₂-y₁)²)

 

Now let's answer the first question by writing the rule for D(x) by plugging in our given coordinates into the distance formula for points (x₁,y₁) and (x₂,y₂):

    (x₁,y₁) = (1,1)

    (x₂,y₂) = (x,f(x)) = (x,ln(x))

    D(x) = √((x-1)² + (ln(x)-1)²)

 

Now let's determine the domain of D(x) - our range of x values for this function.  This will be all values of x which will give real values for y.  We can determine these values by observing these facts in our expression for D(x):

 

    1.  Looking at ln(x), we know that x cannot be 0 because the ln(0) does not exist.

 

    2.  Looking at ln(x), we know that the ln of any negative number will be an imaginary (not real) value.

So x cannot be any negative value.

 

    3.  Looking at the square root function, we know that the square root of any negative number will also be an imaginary (not real) value.  So this is a second reason why x cannot be any negative value. 

 

Therefore, in light of the above facts, we have to conclude that the domain of D(x) is that x > 0.

 

Finally, we'll include the graphs for the function f(x) and a line connecting point (1,1) and any arbitrary point (x,f(x)) on the graph of f(x).  The point we chose was (3,ln(3)) where ln(3) = 1.1 (very close to 1).

 

Below is the clickable link for our graphs:

 

https://dl.dropbox.com/s/zdpij04u3tmmndw/Graph_f%28x%29.png?raw=1

 

Blue curve is the graph of f(x) and green line connects our two red points (1,1) and (3,ln(3)).

 

Thanks for submitting this problem and glad to help.

 

God bless, Jordan.

Answer 2

D(x) = √[(x-1)² + (lnx - 1)²]

 

lnx is undefined if x ≤ 0

 

domain of D(x) is {x l x > 0} = (0, ∞)

Answer 3

Monas, 

 

You cannot use the distance formula on this. This is an arc length formula because In(x) is a curve. The distance formula is used on linear functions. 

 

D(x) = ∫( 1 + (dy/dx)²)^0.5 dx = ∫(1 + (1/x²))^0.5 dx.  If we go from x =1 to x =7 then the distance formula will give you 6.0741,  but the arc length formula will give you 6.3959.

 

This is a big difference.  For example, if you were designing a bridge with an arc, you would be way off.  The iron workers have to wait for the right time of the day, the right temperature, to line up bolts. That would not do them any good in this case because the calculation would be incorrect. 

 

I have not done it yet--because I'm tired and want to go to bed--but I think you can solve that integral with a trig or u substitution.  If you do not know how to do that then leave a comment and I will show you how.  At least I'll try; I'm in the process of moving.