Planar motion (with integrals) | Applications of definite integrals | AP Calculus BC | Khan Academy

A particle moving in the xy-plane has a velocity vector given by (v(t)). It just means that the x component of velocity as a function of time is (\frac{1}{t} + 7), and the y component of velocity as a function of time is (t^4) for time (t \geq 0).

At (t = 1), the particle is at the point (3, 4). So, the first part is: What is the magnitude of the displacement of the particle between time (t = 1) and (t = 3)? We need to figure out its position. We need to round to the nearest tenth.

So, like always, pause this video. I think you will have to use a calculator, but pause this video and try to work through it on your own. We have done questions like this in one dimension, but now we are doing it in two dimensions. The key is to just break it up into the component dimensions.

What we really want to do is find the displacement in the x-direction, so just the change in (x), and then find the displacement in the vertical direction, or the change in (y). We can use those to find the magnitude of the total displacement using the Pythagorean theorem.

Also, if we know the change in (x) and change in (y), we just add the change in (x) to (3) and we add the change in (y) to (4) to find the particle's position at time (t = 3). So, let's figure it out.

The change in (x) from (t = 1) to (t = 3) is the integral of the rate function in the x-direction from time (t = 1) to time (t = 3). In the x-direction, we have (\frac{1}{t} + 7) as our x-velocity as a function of time, so the integral becomes:

[
\int_1^3 \left( \frac{1}{t} + 7 \right) dt
]

What is this going to be equal to? You might want to do (u)-substitution if you're unfamiliar, but you might recognize that the derivative of (t + 7) is just (1). So, you could think of this as (1 \cdot \frac{1}{t + 7}).

We can take the anti-derivative with respect to (t + 7) and get (\ln |t + 7|). We are going to evaluate that at (3) and then subtract it evaluated at (1). So this becomes:

[
\ln |10| - \ln |8| = \ln \left( \frac{10}{8} \right) = \ln(1.25)
]

Let me get my calculator out for a second to calculate that.

Actually, let's just do this in a second. Now, let's figure out our change in (y). Our change in (y) is the integral from (1) to (3). That's our time over which we are thinking about the change. The y-component of our velocity is (t^4):

[
\int_1^3 t^4 dt
]

This will take the reverse power of rule, yielding:

[
\frac{t^5}{5} \bigg|_1^3 = \frac{3^5}{5} - \frac{1^5}{5} = \frac{243}{5} - \frac{1}{5} = \frac{242}{5}
]

That equals (48.4). Let me calculate (\ln(1.25)):

Using my calculator, (\ln(1.25)) is approximately (0.22). So, I figured out our change in (x) and our change in (y.

Actually, we can answer the second part of the question first: What is the particle's position at (t = 3)? At (t = 1), we add each of the components. So:

[
3 + \Delta x \quad \text{and} \quad 4 + \Delta y
]

This will equal (3 + 0.22) and (4 + 48.4):

[
= 3.22 \quad \text{and} \quad 52.4
]

But we still have to answer the first question: What is the magnitude of the displacement? Using the Pythagorean theorem, let's visualize it. Our initial position is at (3, 4).

So, we figured out our change in (x) isn’t much; it's (0.22). Our change in (y) is (48.4). So, we can add them together; if we shift over (0.22) in the x-direction and go for a dramatic change of (48.4) upwards, we can find the hypotenuse.

The magnitude of the displacement is going to be:

[
\sqrt{(\Delta x)^2 + (\Delta y)^2}
]

So we’ll calculate this:

[
\sqrt{(0.22)^2 + (48.4)^2}
]

Let me get my calculator out again for this.

After squaring (0.22) and (48.4), I'll take the square root of that. Our total, the magnitude of our total displacement, is approximately (48.4) when rounded to the nearest tenth.

Now, one thing you might be noting is that it looks like our total displacement (48.4) is the same as our change in (y). The reason it came out this way is because our change in (y) was exactly (48.4) while the magnitude of our displacement was slightly more than (48.4).

But when we round to the nearest tenth, we got (48.4). The reason they’re so close is because our change in (x) was small, at (0.22), while our change in (y) was so significant, resulting in the hypotenuse being only slightly longer than our change in (y).

Thus, in general, you will see that the magnitude of the displacement is larger than the magnitude of either change in (x) or change in (y) alone.