Two questions - (a) Differentiating inverse trig function and (b) vector

[andrew.c]

Struggling with these two questions, any ideas?


A

Homework Statement

Differentiate with respect to x

sin^-1 2x - 4 cos^-1$$\frac{x}{2}$$

Homework Equations

The equations I have in my notes are identical to those derived here...
http://en.wikipedia.org/wiki/Differentiation_of_trigonometric_functions

The Attempt at a Solution

I got

$$\frac{1}{1 - 2x}$$ + $$\frac{8}{2 - x}$$

but I am totally unconvinced by myself!




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B

Homework Statement

Resolve the vector x= (1,4,-3) along and perpendicular to a = (-2,3,1).

Homework Equations

?

The Attempt at a Solution

I don't understand what the question wants me to do...

 

[Chaos2009]

For the first part, $$arcsin(2x)$$, you got the derivative a little wrong.
The derivative of $$arcsin(x)$$ is $$\frac{1}{\sqrt{1 - x^{2}}}$$, but x is actually 2x in this case so replace the x in the denominator of the derivative with 2x, don't forget to square it, and don't forget the chain rule! The derivative of arccos(x) is the opposite of the derivative of arcsin(x), $$\frac{-1}{\sqrt{1 - x^{2}}}$$. You can multiply the -4 through after you find the derivative of arccos(x/2), but don't forget the chain rule here either.

 

[HallsofIvy]

I strongly suspect that, using the rules you refer to, you got
[tex]\frac{1}{\sqrt{1- 4x^2}}[/itex]
and reduce that to 1/(1-2x).

The first is correct, the second is wrong: $\sqrt{a^2- b^2}$ is NOT equal to a- b!

 

[Chaos2009]

$$\sqrt{1 - 4x^{2}} \neq 1 - 2x$$ You just said $$\sqrt{a^{2} - b^{2}} \neq a - b$$ but you went ahead and did it anyway for the first part of the problem.

$$\frac{d}{dx}(arcsin(2x)) = \frac{1}{\sqrt{1 - (2x)^{2}}} * \frac{d}{dx}(2x) = \frac{2}{\sqrt{1 - 4x^{2}}}$$
That is as reduced as it goes unless you want to rationalize the denominator which is unnecessary.

 

[timscully]

For the second question, you need to break the vector down into projection and perpendicular components.

You need to project the vector x on vector a, then take the projection of a away from x. You are then left with the perpendicular component (the same as resolving x and y components of a force vector).

The projection is given by $$\frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a}$$

which gives:

$$\vec{x}.\vec{a} = (1*-2) + (4*3) + (-3*1) = 7$$
$$\left|\vec{a}\right|^{2} = ((-2)^{2}+3^{2}+1^{2}) = 14$$
$$\frac{\vec{x}.\vec{a}}{\left|\vec{a}\right|^{2}}.\vec{a} = (7/14)*(-2,3,1) = (1/2).(-2,3,1)$$

Then take that away from x to get the perpendicular:

$$(1,4,-3) - (1/2)*(-2,3,1) = (1/2) * (4,5,-7)$$

The full vector is then:

$$x = (1/2) * (-2,3,1) + (1/2) * (4,5,-7)$$

Have a look at this:

http://webalg.math.tamu.edu/s03vectors/svec0601.pdf

 

[andrew.c]

Thanks for the help, I eventually did get that last answer, but I did rationalise the denominator.

Ta muchly