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

A particle’s velocity is given by the parametric equations (dπ‘₯/d𝑑) = 2𝑑³ + 5𝑑² βˆ’ 3𝑑 + 1 and (d𝑦/d𝑑) = √(𝑑³ + 3). Find the acceleration vector of the particle at 𝑑 = 2.

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Video Transcript

A particle’s velocity is given by the parametric equations dπ‘₯ by d𝑑 is equal to two 𝑑 cubed plus five 𝑑 squared minus three 𝑑 plus one and d𝑦 by d𝑑 is equal to the square root of 𝑑 cubed plus three. Find the acceleration vector of the particle at 𝑑 is equal to two.

We’re given two parametric equations which represent the velocity of a particle at the time 𝑑. And we want to use these to find the acceleration vector of our particle at the time when 𝑑 is equal to two. We can start by asking the question, what do our parametric equations mean in this case? Well, we know that if π‘₯ of 𝑑 tells us the horizontal position of our particle at the time 𝑑, then dπ‘₯ by d𝑑 is the rate of change of the horizontal position with respect to time. That’s the horizontal velocity of the particle at the time 𝑑. Similarly, if 𝑦 of 𝑑 tells us the vertical position of our particle at the time 𝑑, then d𝑦 by d𝑑 is the rate of change of the vertical position. So d𝑦 by d𝑑 is the vertical velocity of our particle at the time 𝑑.

However, the question wants us to find the acceleration vector of our particle at the time 𝑑 is equal to two. And we recall if we’re given a velocity 𝑣 of 𝑑, then the derivative of 𝑣 with respect to 𝑑 gives us the acceleration. So if we differentiate dπ‘₯ by d𝑑 with respect to time, we get the horizontal acceleration. And if we differentiate d𝑦 by d𝑑 with respect to 𝑑, we get the vertical acceleration. So let’s start by finding the horizontal acceleration of our particle. That’s the derivative with respect to time of the particle’s horizontal velocity. We can differentiate this using the power rule for differentiation. This gives us six 𝑑 squared plus 10𝑑 minus three.

We can then do the same to calculate the vertical acceleration of our particle. We differentiate the vertical velocity with respect to time. We see we have to differentiate the composition of two functions. We’ll do this by using the chain rule. And we recall the chain rule tells us the derivative of 𝑓 evaluated at 𝑔 of 𝑑 is equal to the derivative of 𝑔 multiplied by the derivative of 𝑓 evaluated at 𝑔 of 𝑑. We set our inner function 𝑔 of 𝑑 to be the polynomial 𝑑 cubed plus three and then our function 𝑓 of 𝑑 to be the outer function of the square root of 𝑑. We can differentiate both of these by using the power rule for differentiation. This gives us that 𝑔 prime of 𝑑 is equal to three 𝑑 squared and 𝑓 prime of 𝑑 is equal to one divided by two times root 𝑑.

Substituting these into our chain rule gives us the vertical acceleration is equal to three 𝑑 squared multiplied by one divided by two times the square root of 𝑑 cubed plus three. Which we can simplify to be three 𝑑 squared divided by two times the square root of 𝑑 cubed plus three. So we’ve now found functions for the horizontal and vertical components of the acceleration of our particle. If we substitute 𝑑 is equal to two into these functions, we’ll get the horizontal and vertical components for the acceleration of the particle when 𝑑 is equal to two.

Substituting 𝑑 is equal to two gives us a horizontal acceleration π‘Ž π‘₯ of two is equal to six times two squared plus 10 times two minus three, which is equal to 41. We can do the same to find the vertical acceleration of our particle when 𝑑 is equal to two. It’s three times two squared divided by two times the square root of two cubed plus three. We can calculate this to give us 12 divided by two times the square root of 11. We can simplify this by multiplying both our numerator and our denominator by the square root of 11. We can then calculate this to give us 12 multiplied by the square root of 11 divided by 22. Finally, we can cancel the shared factor of two in our numerator and our denominator to get six times the square root of 11 divided by 11.

So we’ve now found the horizontal and the vertical velocity of our particle when 𝑑 is equal to two. Since we found both the vertical and horizontal components of the acceleration, we can represent this as a vector. This gives us the acceleration of our particle when 𝑑 is equal to two can be represented as 41𝐒 plus six times the square root of 11 divided by 11 𝐣.

Therefore, we’ve shown if a particle’s velocity is given by the parametric equations dπ‘₯ by d𝑑 is equal to two 𝑑 cubed plus five 𝑑 squared minus three 𝑑 plus one and d𝑦 by d𝑑 is equal to the square root of 𝑑 cubed plus three. Then the acceleration vector of the particle at 𝑑 is equal to two is given by 41𝐒 plus six times the square root of 11 divided by 11 𝐣.

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