MHB How Can I Find a Point on sqrt{x} With a Given Slope to (1,1)?

[mathdad]

Find a point P on the curve sqrt{x} such that the slope of the line through P and (1, 1) is 1/4.

There are no sample questions in the book for me to follow.

What are the steps needed for me to solve this problem?

 

[MarkFL]

Let point $P$ be:

$$\left(x,\sqrt{x}\right)$$

Now what is the slope of the line through $P$ and $(1,1)$?

 

[mathdad]

Let point $P$ be:

$$\left(x,\sqrt{x}\right)$$

Now what is the slope of the line through $P$ and $(1,1)$?

I must find the slope between the two points.

 

[MarkFL]

I must find the slope between the two points.

Yes, but what is this slope?

$$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=?$$

 

[mathdad]

Yes, but what is this slope?

$$m=\frac{\Delta y}{\Delta x}=\frac{y_2-y_1}{x_2-x_1}=?$$

m = (sqrt{x} - 1)/(x - 1)

Do you mean to let x = 1?

If so, m = (sqrt{1} - 1)/(1 - 1)

m = 0/0 = 0

 

[MarkFL]

m = (sqrt{x} - 1)/(x - 1)

Yes, we have:

$$m=\frac{\sqrt{x}-1}{x-1}$$

Can you simplify this by writing the denominator as the different of squares and dividing out any common factors?

Do you mean to let x = 1?

If so, m = (sqrt{1} - 1)/(1 - 1)

m = 0/0 = 0

No, we don't want to let $x=1$...otherwise we don't have two distinct points. You would need to use a limit in that case, and you would be computing the instantaneous slope, i.e., the derivative which is a concept from differential calculus. :)

 

[mathdad]

Yes, we have:

$$m=\frac{\sqrt{x}-1}{x-1}$$

Can you simplify this by writing the denominator as the different of squares and dividing out any common factors?
No, we don't want to let $x=1$...otherwise we don't have two distinct points. You would need to use a limit in that case, and you would be computing the instantaneous slope, i.e., the derivative which is a concept from differential calculus. :)

Ok. What is the slope?

 

[MarkFL]

Ok. What is the slope?

You want to first simplify the slope:

$$m=\frac{\sqrt{x}-1}{x-1}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}+1}$$

Now, we are told we want the slope to be 1/4, and so this implies:

$$\sqrt{x}+1=4$$

So, what is the required point?

 

[mathdad]

You want to first simplify the slope:

$$m=\frac{\sqrt{x}-1}{x-1}=\frac{\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}=\frac{1}{\sqrt{x}+1}$$

Now, we are told we want the slope to be 1/4, and so this implies:

$$\sqrt{x}+1=4$$

So, what is the required point?

We equate the slope to 1/4 and solve for x.

 

[mathdad]

1/4 = 1/($$\sqrt{x}$$ + 1)

4 = $$\sqrt{x}$$ + 1

4 - 1 = $$\sqrt{x}$$

3 = $$\sqrt{x}$$

$${3}^{2}$$ = $$({\sqrt{x}})^{2}$$

9 = x

The required point is (9, 3).

 

[MarkFL]

I would suggest looking at the sticky threads in our "Math Formulas (MathJax)" forum, so you can avoid using an awkward mix of text and $\LaTeX$ in your workings. :D

 

[mathdad]

I would suggest looking at the sticky threads in our "Math Formulas (MathJax)" forum, so you can avoid using an awkward mix of text and $\LaTeX$ in your workings. :D

Can you provide the link?

 

[MarkFL]

Can you provide the link?

http://mathhelpboards.com/math-formulas-mathjax-62/

 

[mathdad]

Everytime I try uploading a picture, it tells me the image is too big. What can I do? I would rather upload pictures than deal with LaTex.