find-the-velocity-acceleration-and-speed-of-a-particle-with-the-given-position-function-rt-langle-t2

Question

Find the velocity, acceleration, and speed of a particle with the given position function.

$ r(t) = \langle t^2 + t, t^2 - t, t^3 \rangle $

Foster Wisusik · Accepted Answer

Okay, so this question gives us the position function of a particle, and it wants us to find expressions for the velocity, acceleration and speed off the particle. So for velocity, we just need to take a derivative of the position. And remember, we do this component by component. So the derivative of the first component is to t plus farm. The derivative of the second component is to t minus one, and the derivative of the third component is three t squared. So that&#x27;s our expression for velocity and then for acceleration, we just need to take another derivative. So the derivative of the X component of velocity is too. The derivative of the white component of velocity is too, and the derivative for the Z component of velocity is 60. So now we have expressions for both our velocity and acceleration. But we&#x27;re not done yet because we also need the speed. So speed is equal to the magnitude of velocity. So we&#x27;re just gonna have to find the magnitude of our velocity function. So the magnitude of e of t is equal to Well, we just use our version of the Pythagorean theorem. So again, that&#x27;s just t x t t squared plus d Y t t squared plus d C D t squared And we know each of these components from the last part So give us a big square root to work with here. So our X component is to t plus one are y component is to t minus one and r C component is three t squared. So if we do that, what&#x27;s foil these out and see what we can do Simplifying these so we get for t squared plus 40 plus one plus 40 squared minus 40 plus one men plus nine t to the fourth and now we can see one quick simplification. This cancels with this. So now we get the square root of eight t squared plus two plus nine t to the fourth or registry Raying this in a more standard form the square root of nine T to the fourth plus eight t squared plus two. And this is our final answer

Amy Jiang · Answer

I guess we have our up t. It&#x27;s equal to B squared minus one comma t were asked to find in philosophy. But that&#x27;s this derivative of Artie. So that&#x27;s two t common one. Okay, acceleration is this through the derivative of our velocity? So that&#x27;s just to combat zone and in speed is equal to the magnitude of our velocity, which is square roots. Both 40 squared plus one.

Will Erickson · Answer

either in this problem were asked to find velocity, acceleration and speed if our function are is a position function. So let&#x27;s begin with velocity on. Just remember that velocity is the derivative of position. So we just need our prime of tea and so we can just take the derivatives of each component separately. So the derivative of each of the tea is again. You know the team, the derivative of E to the minus t uh, for use. The train rule will have e to the minus t times the derivative of minus C, which is negative one. So we get minus E to the minus t. So there is our velocity vector. Next, let&#x27;s find acceleration And the acceleration, remember, is just the derivative of velocity. So again we&#x27;ll take the component of the derivative of each component separately. So the derivative eat of the tea once Morrissey to the team, the derivative of minus E to the minus t this time again that either the minus t is its own derivative, except we need to multiply by the derivative of minus t. So we end up with minus minus in front, so we&#x27;ll end up with either tea and then positive e to the minus t could get back to our ah position function in this case now finally were asked to find speed. Let&#x27;s just remember that speed is the length or the norm of our velocity vector. So let&#x27;s look at our first answer and find its length. The length is just the square root of the sum of the squares of the components. So in other words, we have our first complain was even the tea will square that and add our second component squared, which was minus e the minus t square that and now let&#x27;s see what we get. Well, this simplify a little bit. So eat of the T l squared is need of the two t We&#x27;re gonna square the second terms a little end a positive and so we will get. Plus he to the minus two t and that&#x27;s about all we can do. And so we have found speed and we are done

Foster Wisusik · Answer

Okay, so this question wants us to find the velocity, acceleration and speed of a particle with this position function. So to start out, find the velocity which is just the derivative of position. So taking the derivative component wise, we see the derivative of T squared is to t the derivative of to T is too. And the derivative of L N T. Is one over tea and then is for acceleration. We just take the derivative of this velocity well respected time, which is derivative two t&#x27;s two derivative to zero and derivative of one of her tea is negative, one over t squared. And that&#x27;s two out of the three functions that wanted. So now we just need to go about calculating the speed. So the speed is just the magnitude of our velocity which remember we just get by taking the magnitude of each of the components, adding them together and then taken the square. So sincerity found velocity weaken. Just substitute in for each of these. So the x component of velocity is too t. The white component of velocity is just too, and the x component of velocity or sorry, the Z component rather is one over tea. So now we can do some simplification, squaring everything we get four t squared, plus four plus one over t squared. And then if we want to make this easier, we can see that this is actually a perfect square. It&#x27;s actually just to t plus one over tea. Whole thing squared because if you foil it out, we&#x27;ll get that thing under the square root. So this means that our speed simplifies to just to t plus Lord over tea.

Taylor Shimono · Answer

So for this problem, we can go ahead and start by finding the velocity of our particle. I also know that the velocity is just the derivative of our position function. So to find our velocity, all we need to do is take the derivative of T cute. So we&#x27;ll move this three to the front and reduce the export of my one. So this is three t squared, minus the derivative of 90 square. Again, we&#x27;re going to move this tea to the front. So two times nine is 18 times t to the first power and finally plus this 2014. Essentially, we&#x27;re just going to be the one to the front and we just have plus 24. I&#x27;m So now that we have our velocity function, we can find our acceleration function the same way. So we know that since the velocity function is the derivative of the position function, the acceleration function is the second derivative of the position or the first derivative of the velocity. I&#x27;m so essentially we know this is the acceleration function, and all we have to do is differentiate this philosophy function. I&#x27;m so the derivative of three t squared again. We&#x27;ll move that to to the front and reduce our power by one. So that&#x27;s 16 minus. Since we have just a t on the 18 this is just going to be minus 18. And finally, our constant 24 is going to leave us with plus zero at the end. So here we have our velocity function in our acceleration from

Issa Dababneh · Answer

So this video were given the position The position is given by square root of two t I had hospital T j plus feet of the negative TK Hap al or ASA determine the velocities you. So the velocity is the first derivative of position. So we&#x27;re just gonna take the derivative off off with respect. The derivative of square root of two to you with respected teas. Discriminative too. I have says where routed to I had plus the derivative of each of the team and respected is just getting the tea. So plus either the T J hat plus the derivative of either the negative t with respected to use negative to the negatives T. K. So this is our but lost it reality, right? All right, Now the speed is the magnitude of the velocities are gonna determine the magnitude of beat. So we take each i j K component off the velocity and square and then add them together and take this forward of that so it&#x27;s square. So we&#x27;re gonna dig the spirit of two squared must. Either the T squared was negative. Eat the negativity squared, add them together and take a swim. So That&#x27;s just two plus either the duty plus e to the negative. So this negative ghost was worse. Sprayers. All right, now we&#x27;re gonna add these together. Now, we&#x27;re just gonna rearrange things so that we can get rid of this square rooms. We&#x27;re gonna move this either the to t right to the front. And then two, we&#x27;re gonna move it. Right. So we&#x27;re gonna switch these two, and now to we can write it as two times to the tee, times e to the negative. Why? What&#x27;s either the tee times either The negative? This is just either the t plus negative. Well, t plus negative t is zero. So just either zero or just so we just wrote. So either TV crimes, either. The negativity is just a convoluted right. Now, when we look at this, we realize that we can rewrite business each of the t plus e to the negative. T all that square now, why is that? Well, if you look right over here and we actually distribute, so we foil but here either the to t plus either the tee times into the negative G force, either the teeth as either the negativity plus e to the negative to t. And now if we combine like terms so this and this are equal. So if we add them together, we get each of the two t plus each of the tee times you know, the to t plus two times to the tee times e to the negative t bust into the negative tooty. So this is exactly what we had over two square. So they&#x27;re equipped. All right. Now, those square in the spare room cancels or left with either the T plus two are great. So we found the speed. Now the acceleration even think of it as the first derivative of velocity or the second derivative of the position. So now if we take the derivative of square root of two with respect to t well, that&#x27;s just a cause. And so it&#x27;s derivative it zero respect. Now, if we take the derivative off each of the tea with respect to t will, that just stays. Now we think the derivative negative e to the negative t. With respect to you, that&#x27;s just positive. So our acceleration is zero. I have us e to the T J hat plus into the negative T. K. Uh, and that is our excellent

Problem

Find the velocity, acceleration, and speed of a p…

Problem 9 Medium Difficulty

Find the velocity, acceleration, and speed of a particle with the given position function.

$ r(t) = \langle t^2 + t, t^2 - t, t^3 \rangle $

Answer

$\left\langle 2 t+1,2 t-1,3 t^{2}\right\rangle$
$\langle 2,2,6 t\rangle$
$\sqrt{\left(\frac{\partial x}{\partial t}\right)^{2}+\left(\frac{\partial y}{\partial t}\right)^{2}+\left(\frac{\partial z}{\partial t}\right)^{2}}$

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James Stewart

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Lily A.

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$\left\langle 2 t+1,2 t-1,3 t^{2}\right\rangle$$\langle 2,2,6 t\rangle$$\sqrt{\left(\frac{\partial x}{\partial t}\right)^{2}+\left(\frac{\partial y}{\partial t}\right)^{2}+\left(\frac{\partial z}{\partial t}\right)^{2}}$

Calculus

Lily A.

Catherine R.

Heather Z.

Michael J.

Vectors Intro

Vector Basics - Example 1

James StewartCalculus

Lily A.

Catherine R.

Heather Z.

Michael J.

$\left\langle 2 t+1,2 t-1,3 t^{2}\right\rangle$
$\langle 2,2,6 t\rangle$
$\sqrt{\left(\frac{\partial x}{\partial t}\right)^{2}+\left(\frac{\partial y}{\partial t}\right)^{2}+\left(\frac{\partial z}{\partial t}\right)^{2}}$

James Stewart

Calculus