Video: Finding the Position Vector of a Particle given Its Velocity as Two Parametric Equations at a Certain Time

A particle starts its motion at the point 2𝐒 + 4𝐣. The particle’s velocity is given by the parametric equations dπ‘₯/d𝑑 = 4𝑑² +3𝑑 and d𝑦/d𝑑 = 6𝑑³ + 2, where 𝑑 is the time after the particle started its motion. Find the position vector of the particle at time 𝑑.

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Video Transcript

A particle starts its motion at the point two 𝐒, four 𝐣. The particle’s velocity is given by the parametric equations dπ‘₯ by d𝑑 is equal to four 𝑑 squared plus three 𝑑 and d𝑦 by d𝑑 is equal to six 𝑑 cubed plus two, where 𝑑 is the time after the particle started its motion. Find the position vector of the particle at time 𝑑.

The question gives us a pair of parametric equation which tells us the particle’s velocity at the time 𝑑, where 𝑑 is the time after the particle started its motion. The question wants us to use this information to find the position vector of the particle at the time 𝑑. We can start by asking the question, what do our pair of parametric equations mean in this case? We see that we’re given the rate of change of π‘₯ with respect to 𝑑 and the rate of change of 𝑦 with respect to 𝑑. We know that if π‘₯ tells us the horizontal position of our particle at the time 𝑑, then dπ‘₯ by d𝑑, which is the rate of change of the horizontal position with respect to time, must be the horizontal velocity of our particle at the time 𝑑.

Similarly, if 𝑦 is the vertical position of our particle, then d𝑦 by d𝑑 is the rate of change in the vertical position and it’s the vertical velocity of our particle at the time 𝑑. So we need to use the vertical and horizontal velocity functions to find the position vector of the particle at the time 𝑑. So we need something which relates the velocity and the displacement of our particle. We recall if we’re given the displacement function of a particle, then the derivative with respect to time of our displacement gives us the velocity of our particle. And we know that if 𝑠 prime of 𝑑 is equal to 𝑣 of 𝑑, this is the same as saying that 𝑠 of 𝑑 is an antiderivative of 𝑣 of 𝑑.

So we can find our displacement by integrating our velocity function with respect to time. This means if we integrate our horizontal velocity with respect to time, we’ll get a function for the horizontal displacement. And if we integrate the vertical velocity with respect to time, we’ll get a function for the vertical displacement. So we find a function for the horizontal displacement of our particle by integrating the horizontal velocity with respect to time. This gives us the integral of four 𝑑 squared plus three 𝑑 with respect to 𝑑.

And we see we’re integrating a polynomial. We can do this by using the power rule for integration. We add one to our exponent and then divide by this new exponent. This gives us four 𝑑 cubed over three plus three 𝑑 squared over two plus a constant of integration, we will call 𝑐 one. And we can actually solve for the constant of integration in this case. Since our function represents the horizontal displacement at time 𝑑, and we’re told that our particle starts its motion at the point two 𝐒, four 𝐣, which means when 𝑑 is equal to zero, our particle is at the point two 𝐒, four 𝐣. And this tells us that its horizontal displacement is just equal to two.

So if we substitute 𝑑 is equal to zero into our horizontal displacement function, we should get two. Therefore, we have that two is equal to four times zero cubed over three plus three times zero squared over two plus our constant of integration 𝑐 one. We have that four times zero cubed over three is equal to zero and four times zero squared over two is equal to zero. So 𝑐 one is equal to two. So we’ve shown that the horizontal displacement of our particle at the time 𝑑 is given by four-thirds 𝑑 cubed plus three over two 𝑑 squared plus two, which we represent by 𝑠 π‘₯ of 𝑑.

We can then do the same to find a function for the vertical displacement of our particle. We integrate the vertical velocity with respect to time. That’s the integral of six 𝑑 cubed plus two with respect to time, which we can again integrate by using the power rule for integration giving us three 𝑑 to the fourth power over two plus two 𝑑 plus a constant of integration 𝑐 two. And again, we use the fact that the particle starts its motion at the point two 𝐒, four 𝐣. So when 𝑑 is equal to zero, the vertical displacement of our particle is given by four.

So when we substitute 𝑑 is equal to zero into our vertical displacement function, we should get four. This gives us that four is equal to three times zero to the fourth power over two plus two times zero plus 𝑐 two. We have that three times zero to the fourth power over two is just equal to zero and two times zero is equal to zero. So 𝑐 two is equal to four. So we’ve shown that the vertical displacement of our particle at the time 𝑑 is given by three over two 𝑑 to the fourth power plus two 𝑑 plus four which we represent by 𝑠 𝑦 of 𝑑.

Finally, the question wants us to give the position vector of our particle at the time 𝑑. And we can represent the position vector of the particle at the time 𝑑 by using the horizontal and vertical displacement functions. So by using the equations we found for the horizontal and vertical displacement, we can find the position vector of our particle at the time 𝑑.

In conclusion, we’ve shown if a particle started its motion at the point two 𝐒, four 𝐣 and the particle’s velocity was given by the parametric equations dπ‘₯ by d𝑑 is equal to four 𝑑 squared plus three 𝑑 and d𝑦 by d𝑑 is equal to six 𝑑 cubed plus two. Then the position vector of the particle at the time 𝑑 could be given by four-thirds 𝑑 cubed plus three over two 𝑑 squared plus two 𝐒 plus three over two 𝑑 to the fourth power plus two 𝑑 plus four 𝐣.

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