# Math Help - 3cosx and Perpendiculars/Parallel

1. ## 3cosx and Perpendiculars/Parallel

Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.

I found the equation for line PQ:
y-3=(-6/pi)x

And the equation for the line tangent to f at point Q:
y=-3[x-(pi/2)]

But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.

2. Originally Posted by warriors837
Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.

I found the equation for line PQ:
y-3=(-6/pi)x

And the equation for the line tangent to f at point Q:
y=-3[x-(pi/2)]

But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.

I think you did it alright and, imo, you can write $x=Arcsin\frac{2}{\pi} =0.69\,\,\,aprox.$, which indeed is between P and Q on the x-axis

tonio

3. Originally Posted by warriors837
Let f be the function given by f(x)=3cosx. The graph of f crosses the y-axis at point P (0,3) and x-axis at point Q (pi/2,0).

Find the x coordinate of the point of the graph f, between points P and Q, at which the line tangent to the graph of f is parallel to the line PQ.

I found the equation for line PQ:
y-3=(-6/pi)x

And the equation for the line tangent to f at point Q:
y=-3[x-(pi/2)]

But I can't find the x-coordinate (the bolded part), i know that the slope is (-6/pi) and you set that equal to -3sinx, but i got an x-coordinate of sin^(-1)(2/pi). But I am supposed to do it all by hand and I don't think that is a valid answer.
it's a valid solution ...

$x = \arcsin\left(\frac{2}{\pi}\right)$

just doesn't come out "pretty" , that's all.