Give the position vectors of particles moving along various curves in the x y -plane. In each ca

Give the position vectors of particles moving along various...

Key Concepts

Parametric Equations for Curves

Parametric equations are a way to represent curves by expressing the coordinates of the points on the curve as functions of a parameter. This method simplifies the description and analysis of curves, especially when they are not easily defined by a single function of one variable.

Vector-Valued Functions in Kinematics

Vector-valued functions are used to describe the position of a moving particle as a function of time. They combine the individual coordinate functions into a single vector function, allowing for a concise representation of an object's motion in space.

Differentiation of Vector Functions

Differentiating vector-valued functions with respect to time is performed component-wise. The first derivative of the position vector with respect to time yields the velocity vector, and the second derivative yields the acceleration vector. This process is fundamental in analyzing the dynamics of motion.

Velocity and Acceleration Vectors

The velocity vector indicates the rate of change of the position of the particle and its direction of motion, while the acceleration vector represents the rate of change of the velocity. These vectors provide crucial information about how the particle is moving and how its motion is changing over time.

Circular Motion

Circular motion involves movement along a circular path, where the position can be represented using trigonometric functions. In circular motion, the velocity vector is tangent to the path, and the acceleration vector is directed towards the center of the circle (centripetal acceleration), highlighting the relationship between geometric constraints and kinematic quantities.

Transcript

0:00 Hi there.

00:02 In this problem, we are asked to, first of all, find the velocity and acceleration vectors at these two times here, and then we'll sketch them on a curve.

00:16 So why don't we first actually sketch the curve, x squared plus y squared equals 1, which we know is a circle centered at the origin with radius 1.

00:36 And now next, let's find the velocity and acceleration vectors.

00:41 So let's remember velocity is just defined as the derivative of position.

00:47 So we need to find r prime of t.

00:49 I'm going to use angle brackets instead of the ij notation.

00:54 This is a little cleaner.

00:56 So to find the derivative, r prime, we just take the derivative of each component of r separately.

01:02 So the derivative of sine t is cosine.

01:05 T and the derivative of cosine t is minus sine t so there's our velocity vector at any time t and for acceleration we simply remember that acceleration is the derivative of velocity so looking above the derivative of cosine is minus sine t the derivative of minus sine t is minus cosine t so we have formulas for velocity and acceleration now let's actually use these two times so so for time equals pi over 4, the velocity at pi over 4, we just have to look up here.

01:57 The velocity at pi over 4 is going to be cosine of pi over 4 and then minus sign of pi over 4.

02:04 Well, the cosine of pi over 4, remembering our trigonometry, is square root of 2 over 2.

02:12 And that's also the sign of pi over 4.

02:15 So if we just put a negative in front of that, we have our velocity vector at pi over four.

02:23 And for acceleration, so the acceleration vector at pi over four, now we just look at our second answer here for acceleration.

02:33 We'll get the vector.

02:36 Let's see.

02:36 So negative sign of pi over four, we said is negative squared of two over two.

02:41 And negative cosine of pi over four is negative square of two over two.

02:46 Okay, so let's actually at this point, let's draw.

02:51 So we're supposed to draw this, or these two vectors.

02:56 So we have the velocity and acceleration at pi over four.

03:05 First of all, let's find the point on the curve.

03:07 So when t equals pi over four, let's actually look at r, which is position.

03:12 That'll give us what point our positions at on the curve.

03:15 So sine and cosine of pi over 4 squared of 2 over 2, or we can just find the angle of pi over 4, which is 45 degrees.

03:24 Either way, here's our point when t equals pi over 4.

03:29 We're right here on the curve.

03:32 And now velocity, so we'll draw this in green.

03:37 This vector squared to 2 over 2 minus square root of 2 over 2...

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