### Video Transcript

The position of a particle is defined by π₯ is equal to one divided by π‘ plus one and π¦ is equal to three π‘ squared plus seven π‘. Find the speed of the particle when π‘ is equal to three, giving your answer to two decimal places.

Weβre given the position of a particle in terms of a pair of parametric equations. And theyβre given in terms of the variable π‘ which represents the time. The question wants us to use this information to find the speed of the particle when π‘ is equal to three, giving our answer to two decimal places. The question gives us a pair of parametric equations for the position of our particle at the time π‘. It tells us that its π₯-coordinate at the time π‘ is given by one divided by π‘ plus one. And its π¦-coordinate at the time π‘ is given by three π‘ squared plus seven π‘.

Another way of saying this is that our function π₯ of π‘ is equal to one divided by π‘ plus one tells us the horizontal displacement of our particle at the time π‘. And π¦ of π‘ is equal to three π‘ squared plus seven π‘ tells us the vertical displacement of our particle at the time π‘. We need to use this to find the speed of our particle when π‘ is equal to three. And we know that the speed of a particle is the magnitude of the velocity of that particle. And we know that velocity is the rate of change of displacement. So differentiating our horizontal displacement with respect to π‘ gives us the rate of change of horizontal displacement of our particle. And the rate of change of horizontal displacement is just the horizontal velocity of our particle.

Similarly, if we differentiate our vertical displacement function with respect to time, weβll get the vertical velocity of our particle. Then, we can combine the vertical and horizontal components of the velocity to find the velocity of our particle at time π‘. Letβs start by finding the horizontal velocity of our particle at time π‘. Thatβs the derivative of one divided by π‘ plus one with respect to π‘. Weβll differentiate this by using the chain rule. Weβll set π’ equal to π‘ plus one, so our function π₯ of π‘ is equal to one divided by π’. Then the chain rule tells us that dπ₯ by dπ‘ is equal to dπ₯ by dπ’ multiplied by dπ’ by dπ‘.

We see that the derivative of π₯ with respect to π’ is equal to the derivative of one divided by π’ with respect to π’. And by the power rule for differentiation, the derivative of one divided by π’ with respect to π’ is negative one divided by π’ squared. Next, the derivative of π’ with respect to π‘ is equal to the derivative of π‘ plus one with respect to π‘. And the derivative of π‘ plus one with respect to π‘ is just equal to one. Weβll rewrite this expression by using our substitution π’ is equal to π‘ plus one. This gives us that our horizontal velocity is given by negative one divided by π‘ plus one squared. So we found the horizontal component of the velocity of our particle. Letβs now calculate the vertical component of the velocity of our particle.

To do this, we differentiate the vertical displacement of our particle at the time π‘ with respect to time. This gives us the derivative of three π‘ squared plus seven π‘ with respect to time. We can differentiate this using the power rule for differentiation. This gives us that the vertical component of our velocity is given by six π‘ plus seven. Combining the vertical and the horizontal components of our velocity of our particle gives us that the velocity of our particle at the time π‘. Is given by the vector function negative one divided by π‘ plus one squared in the horizontal direction and six π‘ plus seven in the vertical direction.

But remember, the question wants us to find the speed of the particle when π‘ is equal to three. And we know that the magnitude of velocity is equal to the speed. And we know that speed is the magnitude of the velocity. So letβs find the velocity of our particle when π‘ is equal to three by substituting π‘ is equal to three into our function for the velocity. This gives us negative one divided by three plus one squared π’ plus six times three plus seven π£, which we can calculate to give us negative one divided by 16 π’ plus 25π£.

Now, we need to find the magnitude of this vector to find the speed of our particle when π‘ is equal to three. And we know we can calculate the magnitude of a vector π₯π’ plus π¦π£ as the square root of π₯ squared plus π¦ squared. So calculating the magnitude of our velocity when π‘ is equal to three gives us the square root of negative one divided by 16 squared plus 25 squared. Which if we put into our calculator, we get to two decimal places 25.00.

In conclusion, weβve shown if the position of a particle at time π‘ is given by π₯ is equal to one divided by π‘ plus one and π¦ is equal to three π‘ squared plus seven π‘, then the speed of this particle when π‘ is equal to three to two decimal places is 25.00.