Tangential and Normal components of Acceleration | Multi-variable Calculus
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- čas přidán 16. 05. 2020
- Acceleration can be decomposed into a component that is tangential and a component that is normal to the path of a particle, sometimes called centripetal acceleration . In this video, we'll explore the intuition and then use multivariable calculus to come up with formulas to describe these two components, in terms of the velocity of the particle and the curvature of the curve. This creates a right triangle with sides parallel to the unit Tangent vector and the unit Normal vector.
This video follows our previous study of curvature and torsion along the path of a curve: • How curvy is a curve? ...
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woah, I really see the connection between physics and math now. 😳
Also related to Mechanics and Thermodynamics
Simple, easy to remember.
Thanks for not overcooking.
Improved my grasp of orbital mechanics.
At timestamp 6'06, the elegant derivation of a formula met in a course of physics during my last year in high school in ... 1984 and in North Africa;
And I am watching it now in Montreal with a of wind chill of -26 and a curfew, Isn't math beautiful? :-)
And when it is taught in such a brilliant manner, it just becomes poetry;
Kudos to you Professor.
This helped me understand what my Dynamics prof. was talking about. Thank you 🙏🏿
8:57 I think left side should be tangential component of acceleration, regardless, great video!
This is the video I was wondering for 2 years I will be very thankful to you if you make such cool videos on mechanics but can you please make videos on vector calculus in which vector fields are involved.
Thanks a lot sir.. Thia topic is Crystal clear to me now.. It is very helpful..
What a great explanation (seeing from India) I am not a college student I am a high school student your explanation is so good I love it
Thank you very much! Keep making videos like this, easy explanation and not complicated. Thanks!
I don't remember this pedagogical approach during my Uni Physics courses nor mechanical Engineering syllabus !!!! Again much thx
Thanks for simple yet concise explanation!🎉
He teaches everything so well that you feel stupid because you couldn't understand it before.
Thank you, everything's now clear :)
i learned some physics too😁 thank you alot
I'm a mathematician who drives a Porsche 911. On the country roads in the picture, it's definitely the normal component accelleration that is the most fun!
love your animations man. Could you please tell me which software you use to make these mathematical animations
you helped me a lot thank you
Thanks so much.
Incredible! Thank you so much for producing this video. This video helped me understand tangent and normal for my Dynamics class.
Take care. God bless. Jesus loves you.
Beautiful.
Very intuituve, thankss
Great video thank you!
Thanks Sir for previous course OF
GrapH TheorY__🙌
.
.
Love from India__🇮🇳
@@DrTrefor Yup✌️
Lots of, can't express !
Great video. I am having trouble visualizing one thing though: what does the tangential and normal components of gravity look like for a projectile in motion? It seems to me that the rate of change of velocity of such an object would not be 9.8, rather some value less than that (because tangential component, which affects speed, must be less than hypotenuse, which has value 9.8)
The velocity in Horizontal Direction for a projectile is constant, i.e. the acceleration works only in the normal direction. BTW motion of a projectile is not initiated by gravity forces in any case other than free fall.
For purely vertical projectile motion, gravity is either in the same direction, or opposite of the velocity. The acceleration is therefore completely tangential acceleration, since it is always in the same axis as velocity. At the vertex, the projectile is at rest momentarily, and there is acceleration; but that acceleration is neither normal nor tangential, because the velocity is zero, and there is no velocity vector direction that we can use for comparing it.
For projectile motion with a horizontal velocity, the horizontal velocity is constant (neglecting air drag). At the vertex, the acceleration is normal, because acceleration is downward while velocity is purely horizontal. Anywhere else, the acceleration will be a mixture of horizontal and vertical. The farther from the vertex, the more tangential the acceleration will be. The curvature is maximum at the vertex, where the tangential acceleration is zero, and all the acceleration is used for changing its direction.
The rate of change in velocity still is 9.8 m/s^2, even if the rate of change in speed is considerably less than that. For instance, when velocity is 30 degrees above the horizontal, the rate of change in speed (tangential acceleration) is -4.9 m/s^2. But the normal acceleration is 8.49 m/s^2, which adds up in quadrature with the -4.9 m/s^2 to give us a net acceleration of 9.8 m/s^2.
I’m sorry, I have not yet taken Mutlivariable calculus, so this is quite new to me. At around 5 minutes, you used the product rule while differentiating to obtain the acceleration vector, but you described the magnitude of the vector v to be a coefficient. I thought the product rule only applies to the product of two non constant terms. Or in vector calculus does it apply to the magnitude of vectors as well?
@@tobymartiny6784 ahh thank you. In words, the components of the vectors are changing, and since the magnitude is in terms of the components, it changes as well. Thank you my good sir.
@Dr. Trefor Bazett 9:04 the tangential acceleration along the unit vector
Is this not the same tangent vector that appears in the other video? Here, this video, you say that it is “defined” (3:02) as a unit vector. Nothing more than that? I agree, it’s a unit vector, but not just by definition, unless taking the limit in the other video defines it.
I don’t mean to argue with you. I’ve just always approached Calculus with a whole lot of rigor, an approach that has served me well.
I’ve homeschooled my son in (just) math since he was very young. He’s a sophomore now, taking Integral Calculus, and rarely misses. He will take Multivariate next year, which I plan to prep him for over the summer. Getting up to speed now.
We always have to be careful whether we mean "A tangent vector" or "THE unit tangent vector". dr/dt is a generally not unit while dr/ds is a unit and we reserve T for the special case of the unit tangent vector.
I think you mean coefficient aT multiply by vector T at 9:05?
Back in the day of Muscle Cars (circa 1968 +&- 5) we called Normal Acceleration -- lateral G's
sir. how to prove
let p is prime such that p>3 , for all p²-1 is multiple of 24
its satisty. but how to prove formally
thank you so much sir. tnx aloooooooot sir. i wondered by two day how to prove formaly.
tnx alooooooooot sir.
your discrit mathematics videos helped me to pass my mathematics logic nd proof subject.
Tangential to which curvature? Distance vs time graph or simply the trajectory in space?
Tangential to the vector of the instantaneous velocity.
Acceleration vectors are tangent to the hodograph. If you make a space-curve out of the velocity vector vs time, that is called a hodograph. The acceleration vectors are tangent to that.
How do you find out the motivation of all of these mathematical concept. Most of the math text book just give the definition and ask the student to memorized without their physical intepretation ?
Nammude Chacko mashum anghane aayirunnu..a+ b the whole square kaanathe padipichu...avasanam enthaayi - mwon valarnnu Aadu Thoma aayi...
You study other subjects, and recognize the application of what you learn in a math class, to see how they are applied in other subjects.
very nice videos thx alot...i have question please : the intantaneous acceleration is defined by the limit as t goes to zero of the difference of velocity vectors , and this limit ends up in ONE vector only that is either tangentional or normal , i dont get it how there is resultant of two vectors where as ther is only one vector?....thank you in advance
The instantaneous acceleration is a combination of the tangential and normal components.
@@DrTrefor thank you very much for your reply...but i still cant grasp the following : lets say the object is moving on a curve and we apply the definition of instantaneous acceleration : that is the limit of inst. velocity vectors difference as t goes to zero this will produce one vector normal to the instantaneous velocity , so how tangentional vector is produced using the same definition?...thx
@3:18 the magnitude of the velocity vector is a scalar. So how is the product rule applicable?
The magnitude of the velocity is taken to be not constant (in general) throughout the motion. Magnitude of velocity or simply speed is varying with respect to time
In previous classes, we have studied rectilinear motion extensively, (motion along a straight line). in this motion, only the magnitude of velocity can change, not its direction (direction of the motion is along straight line).{v=|v| * t}['v' is 'velocity vector', |v| is it's magnitude and 't' is the unit vector of 'v' (which is constant in this case)] so {d(v)/dt = (d(|v|)/dt )* t} [derivative of a constant times a function is equal to the derivative of the function times the constant, mathematically, d(a * f(x))/dx = a * d(f(x))/dx]
BUT the case of motion we are studying in this video has it's direction changing too. making 't', it's unit vector variable too. and as velocity vector is written to be product of its magnitude and unit vector, the required vector (velocity) is a product of two variable functions (one is magnitude and other is unit vector). thus, to differentiate it, we have to apply product rule for derivative of two multiplied functions
How to relate this to centrifugal force?
The centrifugal "force" is not really a force. It is a consequence of assuming your immediate environment is stationary, when your immediate environment is really accelerating. It is more of an apparent force.
The idea is that when your immediate environment is moving along a curved path, it has to accelerate radially inward to the center of curvature at each point along the way. As a result, if the force causing this motion doesn't uniformly apply to you as well, then you will feel as if there is an apparent force that is radially outward from the center of curvature, as a consequence of your body attempting to follow its default tendency to move in a straight line.
Think of how a car rounds a curve. The traction acts on the tires, and pulls the car to the center of the curve. The car has to transmit this force to your body, but it can only transmit this force via traction at your seat to a local point on your body. Your body then has to transmit this force to the rest of it. Your instinct is to perceive this effect as if there were a force pulling you radially outward, when really, it is the seat traction pulling you radially inward to keep you accelerating with the car.
If gravity is the force causing this motion, like it is for astronauts orbiting Earth in a space station, then you perceive it differently. Gravity acts uniformly on every kilogram of your body, and so does the apparent centrifugal force. As a result, you don't feel any constraint forces in this environment. All of the gravity is "used up" in causing your acceleration. The gravity is still there, it is just nullified as you perceive it. You, the spacecraft, and everything in it, will circle the Earth in sync with one another, and there will be no constraint forces needed to make this happen. You cannot feel true gravity. You feel constraint forces, that you assume are responding to gravity.
Sorry, but I am stuck with the definition of the unit normal vector N(vec) = dT(v)/dt/(|dT/dt|) you say, but In dynamics book this is defined as en = det/du u is the angle and et is the unit tangent vector to the path of the particle. According to you I should define en = det / dt /(|det / dt|)
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But N= dT/ds not dT/dt?
1=√(1×1)=√(-1×-1)=-1
What is wrong in this
Nothing, it's just that 4 function of the square root of c
sqrt c
Is defined as the positive and only the positive solution for x in
x^2 = c
Thus, even though (-1)^2 = 1 = 1×1 and whatever else you listen above to the power of two, we simply choose only positive values for the square root function. What you wrote would be a right solution if you had an equation, but the function of sqrt on it's own has only one value
like what r u doing rn
You look like doctor Strange
I don't have a car