Hey all!
I am currently stuck on this question:
Find the derivative of:
y = arccos (5/x)
I am currently up to:
dy/dx = 5/[(x^2).(sin(arccos (5/x))]
I am unsure how to simplify this.
Any help is awesome!
How are you getting sin(acs(5/x))? I've never seen it done this way before.
Sketch a right triangle with angle $\displaystyle \theta$ adjacent to a side of length 5. The hypotenuse has a length of x. So we know that
$\displaystyle cos(\theta) = \frac{5}{x}$
Thus the sine of this angle is
$\displaystyle sin(acs(5/x)) = \frac{\sqrt{x^2 - 5^2}}{x}$
So your derivative is
$\displaystyle \frac{dy}{dx} = \frac{5}{x \sqrt{x^2 - 25}}$
which is (almost) correct.
A better way to approach this is to use
$\displaystyle \frac{d}{du} acs(u) = -\frac{1}{\sqrt{1 - u^2}}$
So the chain rule says:
$\displaystyle \frac{dy}{dx} = -\frac{1}{\sqrt{1 - \left ( \frac{5}{x} \right )^2}} \cdot -\frac{5}{x^2}$
$\displaystyle \frac{dy}{dx} = \frac{5}{x \sqrt{x^2 - 25}}$
which is again (almost) correct.
Why are neither of these answers quite correct? They work when x is positive. I leave it to you to find out what happens if x is negative. The correct answer is
$\displaystyle \frac{dy}{dx} = \frac{5}{|x| \sqrt{x^2 - 25}}$
-Dan
Thanks for the reply topsquark. However, I think I may be going about this a very difficult way.
The initial question is:
A 5m ladder is resting on a wall. When the base of the ladder is 4m from the wall the base is moving at a rate of 2m/s (away from the wall). Find the rate in which the angle between the ground and the ladder is changing.
Now this is how I get to implicit differentiation. Yet I bet I am making it hard for myself.
Any thoughts on this? Also I'm new to this forum how can I place equations into the post?
If $\displaystyle x$ is the distance of the foot of the ladder from the wall:
$\displaystyle \cos(\theta)=\frac{x}{5}$
so:
$\displaystyle \frac{d}{dt}\cos(\theta)=\frac{1}{5}~\frac{dx}{dt}$
$\displaystyle
-\sin(\theta)~\frac{d\theta}{dt}=\frac{1}{5}~\frac{ dx}{dt}
$
Now you are asked for $\displaystyle \frac{d\theta}{dt}$ when $\displaystyle x=4$ and $\displaystyle \frac{dx}{dt}=2.$
RonL
possible solution:
$\displaystyle
cos\theta = \frac{x}{5}
$
$\displaystyle
\frac{d}{dt} cos(\theta) = \frac{d}{dt} \frac{x}{5}
$
$\displaystyle
-sin(\theta) \frac{d\theta}{dt} = \frac{1}{5} \frac{dx}{dt}
$
$\displaystyle
\therefore \frac{d\theta}{dt} = \frac{-1}{5 sin(\theta)} \frac{dx}{dt}
$
$\displaystyle
cos\theta = \frac{x}{5}
$
$\displaystyle
cos^{2}(\theta) = \frac{x^{2}}{25}
$
$\displaystyle
1 - sin^{2}(\theta) = \frac{x^{2}}{25}
$
$\displaystyle
\therefore sin(\theta) = \frac{\sqrt{25 - x^{2}}}{5}
$
$\displaystyle
\therefore \frac{d\theta}{dt} = \frac{-5}{\sqrt{25 - x^{2}}} \frac{dx}{dt}
$
we know that when $\displaystyle x=4$, $\displaystyle \frac{dx}{dt} = 2$
substituting this gives:
$\displaystyle
\frac{d\theta}{dt} = 2\frac{-5}{\sqrt{25 - 4^{2}}} = \frac{-10}{3}
$
1. Where did this come from.
2. Put brackets around the arguments of trig functions
3. Use \sin \cos etc for the trig functions to avoid them appearing in italic.
4. Consider revising this so it makes sense (say by putting in some punctuation or by some other method make it clear where one equation end and the next starts
5. There is no need to give a complete solution, leave enough for the original poster to complete after showing the key parts of the argument
RonL
We use the mathematical type setting component of LaTeX, a tutorial on it is attached to this thread.
RonL