MHB Equation of slope of a function that passes thru a point

[Nemo1]

Hi Community,

I have this following tutorial question and I am stuck.
View attachment 5516

I understand how to take the derivative and find the tangent line at certain value for $$x$$

Which would say if $$x=2$$ and the derivative of $$\frac{1}{x}$$ equals $$\frac{-1}{x^{2}}$$ then the slope would be $$y=\frac{-1}{4}x$$

I am unsure of how to work out the answer from $$a,\frac{1}{a}$$

Any help would be appreciated in helping me understand this.

Cheers Nemo

 

[MarkFL]

Hello, Nemo! :)

a) To get the equation of a line, all we need is a point on the line and it's slope. We know a point on the tangent line is:

$$\left(a,\frac{1}{a}\right)$$

And we know the slope of the tangent line is:

$$f'(a)=-\frac{1}{a^2}$$

Can you now use the point-slope formula to get the equation of the tangent line?

b) Once you have the equation of the tangent line, try expressing this line in the two-intercept form:

$$\frac{x}{u}+\frac{y}{v}=1$$

Since we know the points:

$$(0,v),\,(u,0)$$

are on the line...and these are the intersects.

c) Use the mid-point formula to show the tangent point is the mid-point of the two intercepts.

 

[Nemo1]

So if I am interpreting this correctly.

Knowing $$\frac{x}{u}+\frac{y}{v}=1$$

I can plugin in $$\left(a,\frac{1}{a}\right)$$

to get $$\frac{a}{1}+\frac{\frac{1}{a}}{2a}=1$$

Giving me the point-slope formula

Where $$1$$ is the derivative of the constant $$a$$ and $$2a$$ is the derivative of $${a}^{2}$$

This seems to have confused me more as I feel like I have skipped a concept in there somehow.

 

[MarkFL]

The point-slope formula is:

$$y=m(x-x_1)+y_1$$

Where $(x_1,y_1)$ is the point we know, and $m$ is the slope we know.

In our case, we know:

$$(x_1,y_1)=\left(a,\frac{1}{a}\right)$$ and $$m=-\frac{1}{a^2}$$

So, plugging in those values, what do you get for the equation of the tangent line?

 

[Nemo1]

The point-slope formula is:

$$y=m(x-x_1)+y_1$$

Where $(x_1,y_1)$ is the point we know, and $m$ is the slope we know.

In our case, we know:

$$(x_1,y_1)=\left(a,\frac{1}{a}\right)$$ and $$m=-\frac{1}{a^2}$$

So, plugging in those values, what do you get for the equation of the tangent line?

Would it be: $$y=-\frac{1}{a^2}(x-a)+\frac{1}{a}$$

I am so unsure at the moment, sorry Mark, I really appreciate your help!

 

[MarkFL]

Would it be: $$y=-\frac{1}{a^2}(x-a)+\frac{1}{a}$$

I am so unsure at the moment, sorry Mark, I really appreciate your help!

Yes, that's correct! (Sun)

I would likely simplify this into the slope-intercept form:

$$y=-\frac{1}{a^2}x+\frac{2}{a}$$

Now, from this you already know the $y$-intercept $$\left(0,\frac{2}{a}\right)$$, so you might just want to let $y=0$ and solve for $x$ to get the $x$-intercept, rather than using the two-intercept form.

Can you now list both intercepts?

 

[MarkFL]

Just to finish up...

The equation of the tangent line in slope-intercept form is:

$$y=-\frac{1}{a^2}x+\frac{2}{a}$$

Arranging the equation into the two-intercept form, we obtain:

$$\frac{x}{2a}+\frac{y}{\dfrac{2}{a}}=1$$

And so we know our intercepts are:

$$\left(2a,0\right),\,\left(0,\frac{2}{a}\right)$$

Now, using the mid-point formula, we find the point midway between these intercepts is:

$$\left(\frac{2a+0}{2},\frac{0+\dfrac{2}{a}}{2}\right)=\left(a,\frac{1}{a}\right)$$

And this is the point of tangency, as required. :)

 

[Nemo1]

Hi Mark,

Thanks for the extra information, I have been revisiting the point-slope and slope-intercept formulas on khan academy.
I was not aware that there was a two-intecerpt form and I am still getting my head around it.

I am wondering now as well that when looking at the graph we can see that one of the intercepts is at $(0,2.5)$ and the other is at $(0,?)$ how does this fit in with the calculated values of $$\left(2a,0\right),\,\left(0,\frac{2}{a}\right)$$?

Should be calculating the actual coordinates of the intercepts with the $x$ $\&$ $y$ $axis$?

I worry that a question like this will be on my exam in 6 weeks time and I'll sit there frozen.

Again, thanks heaps for teaching me these concepts!

Cheers Nemo

 

[MarkFL]

Hi Mark,

Thanks for the extra information, I have been revisiting the point-slope and slope-intercept formulas on khan academy.
I was not aware that there was a two-intecerpt form and I am still getting my head around it.

I am wondering now as well that when looking at the graph we can see that one of the intercepts is at $(0,2.5)$ and the other is at $(0,?)$ how does this fit in with the calculated values of $$\left(2a,0\right),\,\left(0,\frac{2}{a}\right)$$?

Should be calculating the actual coordinates of the intercepts with the $x$ $\&$ $y$ $axis$?

I worry that a question like this will be on my exam in 6 weeks time and I'll sit there frozen.

Again, thanks heaps for teaching me these concepts!

Cheers Nemo

If the $y$-intercept is at $$\left(0,\frac{5}{2}\right)$$ then we an find $a$ from:

$$\frac{2}{a}=\frac{5}{2}\implies a=\frac{4}{5}$$

and so the $x$-intercept is at:

$$(2a,0)=\left(\frac{8}{5},0\right)$$

The graph simply shows you one possible value of $a$...we can let $a$ be any non-zero real number (since this is the domain of $$f(x)=\frac{1}{x}$$).

 

[Nemo1]

Ok, so I have been working thru this last section and have determined the following from Marks help:

First: The slope at the point $$\left(a,\frac{1}{a}\right)$$ is the derivative of the $y$ $coordintate$ $$\frac{1}{a}$$ = $$f'(a)=-\frac{1}{a^2}$$

Second: Using the Point-Slope Formula $$y=m(x-x_1)+y_1$$

We can plugin our slope from step one to get $$y=-\frac{1}{a^2}(x-a)+\frac{1}{a}$$ simplified to $$y=-\frac{1}{a^2}x+\frac{2}{a}$$

Third: Converting $$y=-\frac{1}{a^2}x+\frac{2}{a}$$ into the two-slope form $$\frac{x}{a}+\frac{y}{b}=1$$ we get $$\frac{x}{2a}+\frac{y}{\frac{2}{a}}=1$$

So we know now our intercepts are $$\left(2a,0\right),\,\left(0,\frac{2}{a}\right)$$

Fourth: Using the mid-point formula $$\left(\frac{x_1+x_2}{2},\frac{y_1+y_2}{2}\right)=(x,y)$$

We get $$\left(\frac{2a+0}{2},\frac{0+\dfrac{2}{a}}{2}\right)=\left(a,\frac{1}{a}\right)$$

Fifth: Looking at the graph we know the $y$ $axis$ intercept is at $2.5$ or $$\frac{5}{2}$$ in its fractional form or $$(0,\frac{5}{2})$$ we can plug this into $$\frac{2}{a}=\frac{5}{2}$$ and solve for $a$ and get $$a=\frac{4}{5}$$

Sixth: Now we know that $$a=\frac{4}{5}$$ we can plug this into the original $x$ $intercept$ of $$(2a,0)$$ to get $$(\frac{8}{5},0)$$

Seventh: We now have all we need to determine the Mid-point coordinates:

$$\left(\frac{\frac{8}{5}}{2},\frac{\frac{\frac{2}{4}}{5}}{2}\right)=\frac{4}{5},\frac{5}{4}$$

Graphing this I get the following:
View attachment 5535

Please let me know if I have made an error in my explanation, I find by explaining it back helps me learn and hopefully be able to apply these skills confidently to other new problems.

Cheers Nemo

+ Massive thanks to Mark!

 

[MarkFL]

Hello, Nemo! (Wave)

It's a genuine pleasure to help someone who is truly interested in gaining a better understand. (Sun)

The only thing I would add to in your review of the problem is your first point. When you say you are taking the derivative of the $y$-coordinate, it would probably be better to say you are taking the given function:

$$f(x)=\frac{1}{x}$$

And differentiating w.r.t $x$ to get:

$$f'(x)=-\frac{1}{x^2}$$

and then observing that the slope $m$ of the tangent line passing through $$\left(a,\frac{1}{a}\right)$$ would be:

$$m=f'(a)=-\frac{1}{a^2}$$

While what you did works and is technically correct, I think it is more instructive to look at like this instead. But, we all have our own way of seeing things, and so you should feel free to develop your own style as long as it follows the rules of the calculus.

Nice graph, by the way. (Yes)

 

[Nemo1]

Hi Mark,

I agree about differentiating w.r.t $x$, I feel that because I am only gaining confidence at the moment and at Uni people throw that sentence around a lot it gives me pause in using it.

Cheers Nemo.