# Just another trivial problem

• Apr 15th 2006, 11:32 PM
Jones
Just another trivial problem
Hello, i hope you can help me with this trivial problem(i hope)

Determine the the equation for the tangent to the curve y=tan x/x

at the given point of ((pi/4),(4/pi))

Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

Thank you, Jones
• Apr 15th 2006, 11:57 PM
CaptainBlack
Quote:

Originally Posted by Jones
Hello, i hope you can help me with this trivial problem(i hope)

Determine the the equation for the tangent to the curve y=tan x/x

at the given point of ((pi/4),(4/pi))

Is it just the orinary y-y1=k(x-x1) formulae we need to use or is it different this time?

Thank you, Jones

The equation $\displaystyle y-y_1=k(x-x_1)$ is the equation of the line through
the point $\displaystyle (x_1,y_1)$ with slope $\displaystyle k$. For it to be the equation
of the tangent to the curve at the given point $\displaystyle k$ must be the slope
of the curve at the point. That is:

$\displaystyle k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}$,

and $\displaystyle x_1=pi/4,\ y_1=4/pi$.

RonL
• Apr 16th 2006, 12:32 AM
Jones
Don't we need to differentiate the function first?
• Apr 16th 2006, 12:43 AM
CaptainBlack
Quote:

Originally Posted by Jones
Don't we need to differentiate the function first?

That appears to be what 5th line of my earlier post says :D

RonL
• Apr 16th 2006, 12:51 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
For it to be the equation
of the tangent to the curve at the given point $\displaystyle k$ must be the slope
of the curve at the point. That is:

$\displaystyle k=\frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \} \left \left \vert _{x=\pi /4}$,

$\displaystyle \frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}$.

Therefore

$\displaystyle k=\frac{8}{\pi}-\frac{16}{\pi^2}\approx 0.92534$

RonL
• Apr 25th 2006, 01:58 PM
Jones
Quote:

Originally Posted by CaptainBlack
$\displaystyle -\frac{\tan(x)}{x^2}$.

where did this come from?

Im a little slow you see, almost dumb. therefor you have to be very legible :)
• Apr 25th 2006, 02:08 PM
CaptainBlack
Quote:

Originally Posted by Jones
where did this come from?

Im a little slow you see, almost dumb. therefor you have to be very legible :)

From the product rule for differentiation.

$\displaystyle \frac{d}{dx} f(x)g(x)= f'(x)g(x)+f(x)g'(x)$

and here $\displaystyle g(x)=1/x$ and $\displaystyle f(x)=\tan(x)$ so:

$\displaystyle \frac{d}{dx} \left\{\frac{\tan(x)}{x}\right \}= \frac{1}{x (\cos(x))^2}-\frac{\tan(x)}{x^2}$

I could have used the quotient rule, but since it is equivalent to this
use of the product rule I cant be bothered to remember it (as it is
redundant).

RonL
• Apr 26th 2006, 04:23 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I could have used the quotient rule, but since it is equivalent to this use of the product rule I cant be bothered to remember it (as it is redundant).
RonL

I'm glad I'm not the only one with that "hang-up." :)

-Dan
• Apr 26th 2006, 04:33 AM
CaptainBlack
Quote:

Originally Posted by topsquark
I'm glad I'm not the only one with that "hang-up." :)

-Dan

I'm not sure it is a "hang-up" I find it difficult to see why time and
limited human memory resources are wasted on teaching the quotient
rule.

It seems to me that the teaching of the q-rule in the anomaly -
one of those distrubing features of this reality which makes one suspect
that one has slipped into a parallel universe without having noticed. :eek:

RonL
• Apr 26th 2006, 10:59 AM
topsquark
Quote:

Originally Posted by CaptainBlack
I'm not sure it is a "hang-up" I find it difficult to see why time and
limited human memory resources are wasted on teaching the quotient
rule.

It seems to me that the teaching of the q-rule in the anomaly -
one of those distrubing features of this reality which makes one suspect
that one has slipped into a parallel universe without having noticed. :eek:

RonL

I think I live in a parallel universe. (If I only had a "brane!" sings the scarecrow.)

-Dan