1. ## Problem finding slope

Problem: Find the slope of a straight line that passes through the point (-2, 0) and is tangent to the curve y = sq.root of x.

Okay, so I start by finding the derivative of y which is 1/(2*sq.root of x).

This line is tangent to the curve y = sq.root of x. However, I am uncertain as to how I should proceed in making sure this function passes through (-2, 0). Any help would be greatly appreciated!

2. The given function cannot pass through (-2,0) for the simple reason that it isn't defined there at all.

It is the line whose equation we have to find which passes through the given point.

suppose the line is of the form y=mx+c
m is its slope and c is its y-intercept

As (-2,0) satisfies this line,

0=m(-2)+c
hence c=2m

so the equation is y=mx+2m

In order to find m, we use the fact that the line y=mx-2 is tangent to the curve y=sqrt(x) at some point which is not known to us.
So, for some x,
the slope m= derivative of sqrt(x)
as you've calculated,
m=0.5/sqrt(x) for some particular x.

so 1/sqrt(x) +0.5sqrt(x)=sqrt(x)

0.5sqrt(x)=1/sqrt(x)
so x=2

So m=0.5/sqrt(2)

m=1/2sqrt(2)

hence the equation is y=1/2sqrt(2) + 1/sqrt(2)

x-2sqrt(2)y+2=0

EDITED!

3. If you graph the original equation by squaring both sides to get y^2=x, you'll find that point (1,1) on the graph, will be on a direct path with (-2,0). Then, use those two points to find the slope using the slope formula for a line between two points. Then, use that slope in the y-intercept formula, where you'll have to plug in the (1,1) to find b, just to be doubly correct. I got y=(1/3)x+(2/3).

4. Originally Posted by krje1980
Problem: Find the slope of a straight line that passes through the point (-2, 0) and is tangent to the curve y = sq.root of x.

Okay, so I start by finding the derivative of y which is 1/(2*sq.root of x).

This line is tangent to the curve y = sq.root of x. However, I am uncertain as to how I should proceed in making sure this function passes through (-2, 0). Any help would be greatly appreciated!
See if this makes sense:
Let $y=f(x)$

${f'(x)}=\frac{\sqrt{x}-0}{x-(-2)}$

5. Originally Posted by VonNemo19
See if this makes sense:
Let $y=f(x)$

${f'(x)}=\frac{\sqrt{x}-0}{x-(-2)}$

Er, Lagrange's Theorem?

But the problem is we don't know the point of tangency. And it is the reqd. line and not the function which passes through (-2,0)
The f isn't defined there.
I think my answer is correct.

6. I don't know if this theorem has a name, but I know that this is the best way to do this kind of problem.

$f'(x_0)=\frac{\sqrt{x_0}-0}{x_0-(-2)}$

In words, the above states that the slope of the line tangent to the graph of $f(x)$ at $x=x_0$, will be equal to the slope of a line tangent to the graph of $f(x)$ that passes through the point $(-2,0)$.

Slope equals slope.

Solving for $x_0$, you will get the necessary info to finish the proplem.

7. Bandedkrait,

However, by doing the following I arrived at the correct answer:

I begin, just like you, by definin y = mx + 2m

For m I fill in the derivative of sqrt of x which is 0,5*sqrt of x. For y I fill in sqrt of x. The equation becomes

sqrt x = (0,5sqrt of x)*x + 2*(0,5sqrt of x).

Solving for x we get x = 1/(2*sqrt of 2) which is the right answer according to the book.

Was this the right way to proceed, or did I find the answer just out of sheer luck? I find it somewhat difficult to follow the line of logic, I must admit.

8. Originally Posted by krje1980
Bandedkrait,

However, by doing the following I arrived at the correct answer:

I begin, just like you, by definin y = mx + 2m

For m I fill in the derivative of sqrt of x which is 0,5*sqrt of x. For y I fill in sqrt of x. The equation becomes

sqrt x = (0,5sqrt of x)*x + 2*(0,5sqrt of x).

Solving for x we get x = 1/(2*sqrt of 2) which is the right answer according to the book.
Er I think you've made a mistake here. The slope of the tangent is not 0.5sqrt(x) but 0.5/sqrt(x)

So y=mx+2m

y=sqrt(x)
hence
m=0.5/sqrt(x)

so sqrt(x)= 0.5x/sqrt(x) + 2(0.5/sqrt(x))
hence sqrt(x) =0.5sqrt(x)+1/sqrt(x)
so 0.5sqrt(x)=1/sqrt(x)
hence, we get the value of x as 2

now, use the fact that, m=0.5/sqrt(x)

and find the slope which is 1/{2sqrt(2)}

I don't know if this theorem has a name, but I know that this is the best way to do this kind of problem.

In words, the above states that the slope of the line tangent to the graph of at , will be equal to the slope of a line tangent to the graph of that passes through the point .

Slope equals slope.

Solving for , you will get the necessary info to finish the proplem.
Yes this is the Lagrange's Mean Value Theorem. Ya this could also be used

9. Originally Posted by krje1980
Problem: Find the slope of a straight line that passes through the point (-2, 0) and is tangent to the curve y = sq.root of x.

Okay, so I start by finding the derivative of y which is 1/(2*sq.root of x).

This line is tangent to the curve y = sq.root of x. However, I am uncertain as to how I should proceed in making sure this function passes through (-2, 0). Any help would be greatly appreciated!
The derivative of $y=\sqrt{x}$ is:

$\frac{dy}{dx}=\frac{1}{2\sqrt{x}}$

So the tangent at the point on the curve where $x=x_0$ is:

$y=\frac{1}{2\sqrt{x_0}}x+c$

and when $x=x_0$ we must have $y=\sqrt{x_0}$ so the tangent must be:

$y=\frac{1}{2\sqrt{x_0}}x+\frac{\sqrt{x_0}}{2}$

Now this must pass through (-2,0), so:

$0=-\frac{1}{\sqrt{x_0}}+\frac{\sqrt{x_0}}{2}$

Now solve this for $x_0$ and substitute back into the expression of the slope/derivative.

CB

10. Ya me too did the same.

11. CaptainBlack,

Thank you so much for your suggestion. I follow your logic all the way, except at one point. How is it you get the expression sqrt of x0/2 as the c-value?

12. Originally Posted by krje1980
CaptainBlack,

Thank you so much for your suggestion. I follow your logic all the way, except at one point. How is it you get the expression sqrt of x0/2 as the c-value?
because we have deduced

$m=1/(2(x0)^{1/2})$

and at $x=x0$ ,

${x0}^{1/2} = mx0+c$

Use m and you obtain $c=0.5(x0)^{1/2}$

13. Originally Posted by krje1980
CaptainBlack,

Thank you so much for your suggestion. I follow your logic all the way, except at one point. How is it you get the expression sqrt of x0/2 as the c-value?
This has already been answered but I will repeat it if only because it may be clearer with two explanations.

We have the equation of the line is

$
y=\frac{1}{2\sqrt{x_0}}x+c
$

and we know it must pass through $x=x_0\ y=\sqrt{x_0}$ since this is the point on the curve corresponding to $x=x_0$ where the line is tangent to the curve. So putting these values into the equation of the line we see that:

$\sqrt{x_0}=\frac{1}{2\sqrt{x_0}}x_0+c
$

or:

$
c=\frac{\sqrt{x_0}}{2}
$

CB

14. Great! Thanks