We mathematicians speak another language. But we’re not as bad as physicists!

**Anonymous** asked:

We mathematicians speak another language. But we’re not as bad as physicists!

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**Anonymous** asked:

what is the slope of the line that is tangent to f(x)=ln[arcsinx] at x=sqrt2/ 2?

It is a given result that if f(x) = ln(g(x)), then f’(x) = [g’(x)]/[g(x)].

For our f(x) = ln(arcsinx), we have f’(x) = [d(arcsinx)/dx] * [1/arcsinx]

It is also a general result that [d(arcsinx)/dx] = 1/root(1-x^2)

So now we have that f’(x) = 1/[root(1-x^2)*arcsinx], at which point you can just substitute in the value of x to be root(2)/2/.

Hope this has helped, have a nice day.

**Anonymous** asked:

how do i find the slope of the line normal to the graph of r=2cos(theta/2) at theta=pi

I assume that by slope you mean the gradient of the graph.

First off, the normal is the line that is perpendicular to the tangent, and it is a given result that for two perpendicular lines with gradients m, n, then mn = -1.

Therefore, if the gradient of the tangent of the graph at some point is a, then the gradient of the normal is -1/a. However, in the case of a = 0, then the normal is a vertical line and thus has an “infinite” gradient.

Therefore all you need to do is differentiate the graph function you have (work out dr/dtheta), substitute in the value of theta, and then divide -1 by this value.

r = 2cos(theta/2) => dr/dtheta = 2(1/2)(-sin(theta/2)) = -sin(theta/2)

Theta = pi => dr/dtheta = -sin(pi/2) = -1. Therefore, the gradient of the normal is (-1)/(-1) = 1, and so the equation of the normal is of the form y = x + c, where c is a constant to be found.

Hope this has helped, have a nice day.

**Anonymous** asked:

how do i find the lim of x(1-tanx) / cosx - sinx as x approaches pi/4 ?then how do i use that result to derive an approximation of secx in terms of x for values of x near 0?

If you take x(1-tanx)/(cosx-sinx) and rewrite it, you get (x(cosx-sinx)/(cosx))/(cosx-sinx) = x/cosx

Taking x = pi/4, we get that x/cosx = (pi/4)/(1/root(2)) = root(2)*pi/4

As for the second half of your question I have to admit that I don’t understand what you’re talking about so could you please explain to me what is being asked? I’m sorry for the inconvenience.

My understanding is that you want to estimate secx as x tends to 0, but that’s strange since secx = 1 when x tends to 0. If you are however talking about how to use the original expression, divide whatever value you get by the x-value and that’ll give the value of secx.

A third meaning I thought your question may have is that it wants you to use the small angle approximations, namely:

sinx ~ x

tanx ~ x

cosx ~ 1 - (x^2)/2

These approximations are only true for values of x that are very small and for x is in radians.

Using these, we see that x(1-tanx)/(cosx-sinx) ~ x(1-x)/(1-[(x^2)/2]-x) = 2x(1-x)/(2-2x-x^2), and dividing through by x, since the original expression evaluated to x/cosx = xsecx, we get 2(1-x)/(2-2x-x^2) which can then be used to approximate secx for small values of x.

Hope this has been helpful, and if I interpreted the question wrong please let me know. Have a nice day.

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