MHB Differentiation and Tangent Lines.

[tc903]

Given$$ f(x) = \arctan\left({\frac{\sqrt{1+x}}{\sqrt{1-x}}}\right)$$

I differentiated and this was my answer.

$$\d{y}{x} = \frac{1}{2\sqrt{1+x}\sqrt{1-x}{(1-x)}^{2}}$$

I used implicit differentiation on the elliptic curve $${x}^{2}+4{y}^{2} = 36$$ and it wants two horizontal tangents through $$(12,3)$$

Finding the derivative implicitly I get.

$$\d{y}{x} = \frac{-x}{4y}$$

 

[MarkFL]

I wrapped your $\LaTeX$ coding with $$$$ tags, so they would parse correctly. :D

1.) W|A returns a different result for $$\d{f}{x}$$

2.) You have implicitly differentiated correctly. But, there would only be one horizontal tangent through the given point. I would actually work this problem without using the calculus. I would let the tangent lines be:

$$y=m(x-12)+3$$

And then substitute for $y$ into the equation of the ellipse, and then for the resulting quadratic in $x$, require the discriminant to be zero, resulting in a quadratic in $m$, which you can solve, and then you have the two required tangent lines.

 

[tc903]

I found $$\d{f}{x}$$ for the first one both using quotient rule and product rule. I changed the latex so it showed my answer now.

 

[MarkFL]

I found $$\d{f}{x}$$ for the first one both using quotient rule and product rule. I changed the latex so it showed my answer now.

Without seeing your work, I can't tell where your error is. Here is what I get:

I would choose to write:

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

Hence:

$$\d{f}{x}=\frac{1}{\left(\sqrt{\dfrac{1+x}{1-x}}\right)^2+1}\cdot\frac{1}{2}\left(\frac{1+x}{1-x}\right)^{-\frac{1}{2}}\cdot\frac{(1-x)(1)-(1+x)(-1)}{(1-x)^2}$$

$$\d{f}{x}=\frac{1-x}{2}\cdot\frac{1}{2}\left(\frac{1-x}{1+x}\right)^{\frac{1}{2}}\cdot\frac{2}{(1-x)^2}=\frac{1}{2\sqrt{1-x^2}}$$

This agrees with the result I obtained using W|A last night. :D

 

[tc903]

I know what I missed. Thanks MarkFL, I will get back on the second problem soon.

 

[tc903]

Alright, here is my work.

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

$$ \d{f}{x} = \frac{1}{1+{\left(\frac{\sqrt{1+x}}{\sqrt{1-x}}\right)}^{2}}\left[\d{}{x}{\left(\frac{1+x}{1-x}\right)}^{1/2}\right] [/MATh]

$$ \d{f}{x} = \frac{1}{\frac{1-x}{1-x}+\frac{1+x}{1-x}}\left[(\frac{1}{2}\left({\frac{1+x}{1-x}}\right)^{\frac{-1}{2}})\d{}{x}\left(\frac{1+x}{1-x}\right)\right] [/MATh]

$$ \d{f}{x} = \frac{1}{\frac{2}{1-x}}\left[\frac{\sqrt{1-x}}{2\sqrt{1+x}}\left(\frac{\left(1-x\right)\d{}{x}\left(1+x\right)-\left(1+x\right)\d{}{x}\left(1-x\right)}{{\left(1-x\right)}^{2}}\right)\right] $$

$$ \d{f}{x} = \frac{1-x}{2}\left[\frac{\sqrt{1-x}}{2\sqrt{1+x}}\left(\frac{\left(1-x\right)-\left(1+x\right)\left(-1\right)}{{\left(1-x\right)}^{2}}\right)\right] $$

$$ \d{f}{x} = \frac{1-x}{2}\left[\frac{\sqrt{1-x}}{2\left(\sqrt{1+x}\right)}\left(\frac{\left(1-x\right)+\left(1+x\right)}{{\left(1-x\right)}^{2}}\right)\right] $$

$$ \d{f}{x} = \frac{1-x}{2}\left[\frac{\sqrt{1-x}}{2\sqrt{1+x}}\left(\frac{2}{{\left(1-x\right)}^{2}}\right)\right] $$

$$ \d{f}{x} = \frac{1-x}{2}\left[\frac{\sqrt{1-x}}{\sqrt{1+x}{\left(1-x\right)}^{2}}\right] $$

$$ \d{f}{x} = \d{y}{x} = \frac{\sqrt{1-x}}{2\sqrt{1+x}\left(1-x\right)} $$

I know that is wrong.

 

[MarkFL]

Your result is correct. What you have can be simplified a bit and be written in the same form I gave. :D

 

[tc903]

The second problem.

$$ y - {y}_{1} = m\left(x-{x}_{1}\right) $$

$$ y - 3 = m\left(x-12\right) $$

$$ y = m\left(x-12\right) + 3 $$Substituting the above,$$ {x}^{2} + 4{y}^{2} = 36 $$

$$ {x}^{2} +4\left(m\left(x-12\right)+3\right)^{2} = 36 $$

$$ {x}^{2} + 4\left(mx-12m+3\right)^{2} = 36 $$

 

[MarkFL]

Good...now what you want to do is express this quadratic in $x$ in standard form:

$$ax^2+bc+c=0$$

And then set the discriminant to zero...can you explain why we need the discriminant to be zero?

 

[tc903]

I had to leave.

Did you mean $$ a{x}^{2}+bx+c = 0 $$

$$ {b}^{2} - 4ac = 0 $$ So that I may find repeated real solutions.

$$ {x}^{2} + 4\left({(mx)}^{2}-12{m}^{2}x+3mx-12{m}^{2}+144{m}^{2}-36m+3mx-36m+9\right) = 36 $$

Skipping steps shown.

$$ {x}^{2}+4{(mx)}^{2}-96{m}^{2}x+24mx+576{m}^{2}-288m = 0 $$

 

[MarkFL]

That's correct, and then putting it in standard form, I get:

$$\left(4m^2+1\right)x^2+24m(1-4m)x+288m(2m-1)=0$$

Now, the reason we want the discriminant to be zero is as you said, we want only 1 repeated root, which corresponds to the case of the line being tangent to the ellipse, i.e., there is only 1 point of intersection. So, equating the discriminant to zero, what do you find?

 

[MarkFL]

If we set the discriminant to zero, we obtain:

$$\left(24m(1-4m)\right)^2-4\left(4m^2+1\right)\left(288m(2m-1)\right)=0$$

This reduces to:

$$576m(3m-2)=0$$

Hence, we find:

$$m\in\left\{0,\frac{2}{3}\right\}$$

And so our two tangent lines are:

$$y_1=0(x-12)+3=3$$

$$y_2=\frac{2}{3}(x-12)+3=\frac{2}{3}x-5$$

Now, you may be thinking, "but I am taking calculus...shouldn't I be using the derivative I found?"

So, I will outline what I would do to use the calculus here:

1.) Label the tangent point as $(x,y)$.

2.) Equate your derivative with the slope of the line passing through the tangent point and the given point $(12,3)$, to get a relationship between $x$ and $y$.

3.) Use the equation of the ellipse to greatly simplify this relationship, to obtain a linear relationship between $x$ and $y$.

4.) Substitute for either $x$ or $y$ using the linear equation for step 3.) into the equation of the ellipse to obtain a quadratic, which you then solve.

5.) You should now have two points, which you substitute into your derivative to get the slopes of the two tangent lines.

6.) Use the point-slope formula to determine the two tangent lines.

Can you post your working using this method?

 

[MarkFL]

Okay, I going to wrap this up for other readers. :D

2.) Equate your derivative with the slope of the line passing through the tangent point and the given point $(12,3)$, to get a relationship between $x$ and $y$.

$$-\frac{x}{4y}=\frac{3-y}{12-x}-\implies x^2+4y^2-12(x+y)=0$$

3.) Use the equation of the ellipse to greatly simplify this relationship, to obtain a linear relationship between $x$ and $y$.

Since we know $x^2+4y^2=36$, we obtain:

$$36-12(x+y)=0\implies y=3-x$$

4.) Substitute for either $x$ or $y$ using the linear equation for step 3.) into the equation of the ellipse to obtain a quadratic, which you then solve.

$$x^2+4(3-x)^2=36\implies x(5x-24)=0$$

5.) You should now have two points, which you substitute into your derivative to get the slopes of the two tangent lines.

$$(x,y)=(0,3),\,\left(\frac{24}{5},-\frac{9}{5}\right)$$

$$\left.\d{y}{x}\right|_{(x,y)=(0,3)}=-\frac{0}{4(4)}=0$$

$$\left.\d{y}{x}\right|_{(x,y)=\left(\frac{24}{5},-\frac{9}{5}\right)}=-\frac{\frac{24}{5}}{4\left(-\frac{9}{5}\right)}=\frac{2}{3}$$

6.) Use the point-slope formula to determine the two tangent lines.

$$y_1=0(x-12)+3=3$$

$$y_2=\frac{2}{3}(x-12)+3=\frac{2}{3}x-5$$

 

[tc903]

Thanks MarkFl, you were extremely helpful and our answers agree.