# Calculus gradient at a specific time

1. Oct 16, 2011

### e to the i pi

The path of a particle is given by r(t) = tsin(t) * i - tcos(t) * j where t≥0. The particle leaves the origin at t = 0 and then spirals outwards.

Let θ be the acute angle at which the path of the particle crosses the x-axis.
Find tan(θ) when t = 3pi/2.

I was able to figure out a few things:
At t = 3pi/2, the particle crosses the x-axis for a second time (first being at t = pi/2).
At t = 3pi/2, the speed of the particle is (1/2)sqrt(9*pi^2 + 4).
Tan(θ) = gradient = dy/dx
Other than that, I don't know where to start or what to do.

2. Oct 16, 2011

### HallsofIvy

Staff Emeritus
Why are you talking about the second time the particle crosses the x-axis? The problem you give says nothing about "second time". Also I see no mention of "speed" so I don't see why you are calculating that.

dy/dx= (dy/dt)/(dx/dt). Can you calculate that? Remember that when the particle crosses the x-axis, cos(t)= 0.

3. Oct 16, 2011

### e to the i pi

I calculated that:
v(t) = (sin(t) + tcos(t))*i + (tsin(t) - cos(t))*j
But now I am completely lost. This is the graph of velocity vs time, not position vs time. If I substituted in t = 3pi/2, I get:
v(3pi/2) = -1*i + (-3pi/2)*j
Which doesn't make any sense to me because that is a velocity, not an angle.

4. Oct 16, 2011

### HallsofIvy

Staff Emeritus
I told you exactly what to do in my first post. Read it.

5. Oct 16, 2011

### e to the i pi

So if I let x = i and y = j, then:
dx/dt
= i component of r'(t)
= sin(t) + tcos(t)
dy/dt
= j component of r'(t)
= tsin(t) - cos(t)
(dy/dt)/(dx/dt)
= (tsin(t) - cos(t)) / (sin(t) + tcos(t))
Let cos(t) = 0
= tsin(t) / sin(t)
= t
since t = 3pi/2
tan(θ) = 3pi/2

Is that the answer?