Math Help - Derivative Question

1. Derivative Question

A line passes through the point (-root3, -pi/6) and is tangent to the graph y=sinx at a point for which x is an element of [0. pi/2] detirmine the slope of the tangent.

root is a square root,

I really dont know where to go with this
any help would be great

2. Originally Posted by tribar
A line passes through the point (-root3, -pi/6) and is tangent to the graph y=sinx at a point for which x is an element of [0. pi/2] detirmine the slope of the tangent.

root is a square root,

I really dont know where to go with this
any help would be great

let "a" be the point in $[0,\frac{\pi}{2}]$

then the point $(a,f(a))=(a,\sin(a))$ must also be on the tangent line.

so we can now compute the slope using these 2 points

$m=\frac{\sin(a)+\frac{\pi}{6}}{a+\sqrt{3}}$

but we also know that the slope must be equal to the derivative evaluated at a so we get

$f'(x)=\cos(x) \to m=f'(a)=\cos(a)$

We now have a system of equations in "m" and "a"

setting them equal gives

$\frac{\sin(a)+\frac{\pi}{6}}{a+\sqrt{3}}=\cos(a)$

as far as I know (any input we be awsome )

this doesn't have a closed form solution(there may be another way )

but I solved the equation numericlly with Maple and it gives

$a \approx 1.047197551$ this is in radians.

I hope this helps.

3. Thanks
thats where i was going with the eqn, and i got to where you are....

the overall answer is supposed to be (for the eqn of the line)
y= 1/2x -pi/6 + (root3/2)

thanks again!

4. Originally Posted by tribar
Thanks
thats where i was going with the eqn, and i got to where you are....

the overall answer is supposed to be (for the eqn of the line)
y= 1/2x -pi/6 + (root3/2)

thanks again!

Numerically we have the right slope

$\frac{\pi}{3} \approx 1.047197551$

but how they sovled without using a numerical approximation I have no Idea.

5. i see what you mean, my trig/algebra skills arnt as good as they used to be

but ill try to work through it, thus far i have it at
pi/6 = (x + sqrt3)cosx - (1/cosx)