In this video, I go over further into vector functions, and this time look at motion in space in terms of the velocity, acceleration, and position of objects moving in 2D or 3D space. Using the fact that the rate of change of the position of an object is the velocity, and likewise that the rate of change of velocity is the acceleration, I show we can write the velocity and acceleration of an object just in terms of the derivatives of the position vector. Likewise, going backwards via vector integrals, we can obtain the velocity from the acceleration vector, and the position from the velocity vector. I also do an example on centripetal force, which is derived from Isaac Newton's 2nd law of motion, F = m⋅a, and arises for circular motion. In a future video, I will show how these equations are used in the famous Kepler's Laws of planetary motion!
#math #vectors #calculus #space #physics
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In this section we show how the ideas of tangent and normal vectors and curvature can be used in physics to study the motion of an object, including its velocity and acceleration, along a space curve.
In particular, we follow in the footsteps of Newton by using these methods to derive Kepler’s First Law of planetary motion.
Suppose a particle moves through space so that its position vector at time t is r(t).
Notice from the figure above that, for small values of h, the vector
approximates the direction of the particle moving along the curve r(t).
Its magnitude measures the size of the displacement vector per unit time.
The above vector gives the average velocity over a time interval of length h and its limit is the velocity vector v(t) at time t:
Thus the velocity vector is also the tangent vector and points in the direction of the tangent line.
The speed of the particle at time t is the magnitude of the velocity vector, that is, |v(t)|.
This is appropriate because, from the above equation, and from the equation from my earlier video on the arc length function, we have:
As in the case of one-dimensional motion, the acceleration of the particle is defined as the derivative of the velocity:
The position vector of an object moving in a plane is given by r(t) = t3 i + t2 j.
Find its velocity, speed, and acceleration when t = 1 and illustrate geometrically.
Solution:
The velocity and acceleration at time t are:
and the speed is obtained from the formula for the length or magnitude of a vector, which I showed in my earlier video:
When t = 1, we have:
These velocity and acceleration vectors are shown in the figure below.
Let's also graph in the GeoGebra 2D graphing calculator:
Find the velocity, acceleration, and speed of a particle with position vector:
Solution:
The figure below shows the path of the particle with the velocity and acceleration vectors when t = 1.
Let's also graph in the GeoGebra 3D graphing calculator:
The vector integrals that were introduced in my earlier video can be used to find position vectors when velocity or acceleration vectors are known, as in the next example.
A moving particle starts at an initial position r(0) = < 1, 0, 0> with an initial velocity of v(0) = i - j + k.
Its acceleration is a(t) = 4t i + 6t j + k.
Find its velocity and position at time t.
Solution:
To determine the value of the constant vector C, we plug in our given value for v(0).
Likewise, using our given value for r(0) to solve for the integration constant vector.
The expression for r(t) that we obtained in Example 3 was used to plot the path of the particle in the figure below for 0 ≤ t ≤ 3.
Let's also graph in the GeoGebra 3D graphing calculator:
In general, vector integrals allow us to recover velocity when acceleration is known and position when velocity is known:
If the force that acts on a particle is known, then the acceleration can be found from Newton’s Second Law of Motion.
The vector version of this law states that if, at any time t, a force F(t) acts on an object of mass m producing an acceleration a(t) , then
An object with mass m that moves in a circular path with constant angular speed ω has position vector r(t) = a cos ωt i + a sin ωt j.
Find the force acting on the object and show that it is directed toward the origin.
Solution:
To find the force, we first need to know the acceleration:
Therefore Newton’s Second Law gives the force as
This shows that the force acts in the direction opposite to the radius vector r(t) and therefore points toward the origin (see the figure below).
Such a force is called a centripetal (center-seeking) force.
The angular speed of the object moving with position P is ω = dθ / dt, where θ is the angle shown below.
Let's also graph in the GeoGebra 2D graphing calculator:
A projectile is fired with angle of elevation α and initial velocity v0 as in the figure below.
Assuming that air resistance is negligible and the only external force is due to gravity, find the position function r(t) of the projectile.
What value of α maximizes the range (the horizontal distance traveled)?
Solution:
We set up the axes so that the projectile starts at the origin.
Since the force due to gravity acts downward, we have:
Integrating again, we obtain:
Thus the position vector of the projectile is given by:
If we write |v0| = v0 (the initial speed of the projectile), then:
Thus the position vector of the projectile becomes:
The parametric equations of the trajectory are therefore:
If you eliminate t from the above equations, you will see that y is a quadratic function of x.
So the path of the projectile is part of a parabola.
The horizontal distance d is the value of x when y = 0.
Setting y = 0, we obtain:
This second value of t then gives:
Recall the sine double angle identity from my earlier video.
Clearly, d has its maximum value when sin 2α = 1, that is, α = π/4.
A projectile is fired with muzzle speed 150 m/s and angle of elevation 45° from a position 10 m above ground level.
Where does the projectile hit the ground, and with what speed?
Solution:
If we place the origin at ground level, then the initial position of the projectile has coordinates (0, 10) and so we need to adjust the earlier parametric equations of trajectory by adding 10 to the expression for y.
Impact with the ground occurs when y = 0, which we can solve via the quadratic formula from my earlier video, and using only the positive value of t, we get:
So, the projectile hits the ground about 2,306 m away.
Calculation check:
75^2 * 2 = 11,250
4 * 4.9 * 10 = 196
11,250 + 196 = 11,446
Sqrt(11,446) = 106.9859803899558
75 * sqrt(2) = 106.0660171779821
(75 * sqrt(2) + sqrt(11,446))/9.8 = 21.739999751830475 * sqrt(2) * 21.74 = 2,305.875213449331
The velocity of the projectile is:
So its speed at impact is:
Calculation Check:
Sqrt((75 * sqrt(2))^2 + (75*sqrt(2) - 9.8 * 21.74)^2) = 150.6519184092692
When we study the motion of a particle, it is often useful to resolve the acceleration into two components, one in the direction of the tangent and the other in the direction of the normal.
Recall the Unit Tangent, Unit Normal, and Unit Binormal vectors from my earlier video.
If we write v = |v| for the speed of the particle, then the unit tangent vector T is:
If we differentiate both sides of this equation with respect to t, we get:
If we use the expression for the curvature given by my earlier video, then we have:
The unit normal vector was defined as N = T'/|T'|, so we have:
Writing aT and aN for the tangential and normal components of acceleration, we have:
This resolution is illustrated in the figure below.
Let’s look at what the acceleration vector components formula says.
The first thing to notice is that the binormal vector B is absent.
No matter how an object moves through space, its acceleration always lies in the plane of T and N (the osculating plane, which I covered in my earlier video).
Recall that T gives the direction of motion and N points in the direction the curve is turning.
Next we notice that the tangential component of acceleration is v', the rate of change of speed, and the normal component of acceleration κv2, the curvature times the square of the speed.
This makes sense if we think of a passenger in a car—a sharp turn in a road means a large value of the curvature, so the component of the acceleration perpendicular to the motion is large and the passenger is thrown against a car door.
High speed around the turn has the same effect; in fact, if you double your speed, aN is increased by a factor of 4.
Although we have expressions for the tangential and normal components of acceleration, it’s desirable to have expressions that depend only on r, r', r''.
To do this end we take the dot product of v = vT with a as given by the acceleration vector components equation, and simplify using the geometric equation of the dot product from my earlier video, as well as the properties of the dot product.
Using the formula for curvature given by the earlier video thumbnail above, we have:
A particle moves with position function r(t) = <t2, t2, t3>.
Find the tangential and normal components of acceleration.
Solution:
Therefore the earlier formula for the tangential component of acceleration gives:
The equation for the normal component of acceleration gives:
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