Question

Consider an object moving along the curve y=sin(x), starting at the origin. Assuming that the object's x-coordinate, s, is increasing at a rate of 3 ft/sec, what is the rate of change of p at the moment when s=pi/4? (P denotes the x-intercept of the line tangent to the curve y=sin(x) at the point (s, sin(s))). Is p at that moment moving towards or away from the origin?

Answer 1

do you have the equation of the line tangent ?

Answer 2

There is no equation given.

Answer 3

y = sin (x)
dy/dx = cos (x)

tangent line at (s, sin(s))
y - sin (s) = dy/dx (x - s)
y - sin (s) = cos (x) (x - s)

x-intercept is at (p, 0):
0 - sin (s) = cos (p)(p - s)
-sin (s) = cos (p) (p - s)

then take the derivative to solve for dp/dt
-cos(s) ds/dt = cos (p) (dp/dt - ds/dt) - (p - s)sin (p)(dp/dt)

I think...

Answer 4

@peachpi

you're kinda kinda where i come out but i can't totally reconcile our separate approaches

we have position vector \(\vec r = <s, \sin s>\)

and so velocity vector \(\vec r' = \vec v = <\dot s, \cos s ~ \dot s>\)


The velocity vector is the tangent vector so the tangent line is parameterised as

\(\vec l = <s, \sin s> + \lambda <\dot s, \cos s ~ \dot s>\)


This will cross the x axis at

\(y = \sin s + \lambda \cos s \dot s = 0\)

\(\implies \lambda = - \dfrac{\sin s}{\cos s ~ \dot s} = - \dfrac{\tan s}{ \dot s}\)

so x value, ie p, is

\(p = s + \lambda \dot s = s - \tan s\)

At that point \(\dot p = \dot s - \sec^2 s ~ \dot s\)

\(\dot p = - \dot s \tan^2 s\)

reality check....does this make sense at \(s = 0, \frac{\pi}{2}\) etc. and it kinda does

so plug in \(s = \frac{\pi}{4}, \dot s = 3\)