# Question Video: Finding the Time When a Moving Particle on a Curve Defined by a Pair of Parametric Equations Is Moving with a Certain Acceleration Mathematics • Higher Education

If a particle is moving on a curve defined by the parametric equations d𝑥/d𝑡 = 𝑡² − 6𝑡 and d𝑦/d𝑡 = 𝑡² + 5𝑡 − 3, find the time, to the nearest tenth, at which 𝑎 = 64.

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### Video Transcript

If a particle is moving on a curve defined by the parametric equations d𝑥 by d𝑡 is equal to 𝑡 squared minus six 𝑡 and d𝑦 by d𝑡 is equal to 𝑡 squared plus five 𝑡 minus three, find the time, to the nearest tenth, at which 𝑎 is equal to 64.

We’re told that a particle is moving on a curve which is defined by a pair of parametric equations, which we’re given as d𝑥 by d𝑡 is equal to 𝑡 squared minus six 𝑡 and d𝑦 by d𝑡 is equal to 𝑡 squared plus five 𝑡 minus three. The question wants us to find the time 𝑡 to the nearest tenth at which 𝑎 is equal to 64. That’s the acceleration of our particle. And we know that we can find the acceleration of a particle by using the velocity of the particle. The acceleration is equal to the rate of change of velocity with respect to time.

And we can see in our question that we’re given the rate of change of 𝑥 with respect to time and the rate of change of 𝑦 with respect to time. And we know that 𝑥 represents the horizontal position of a particle at the time 𝑡. So d𝑥 by d𝑡, which is the rate of change of the horizontal position with respect to time, must be the horizontal velocity of our particle. Similarly, we know that 𝑦 is the vertical position of our particle at the time 𝑡. So d𝑦 by d𝑡 must be the vertical velocity of our particle at the time 𝑡.

Then we notice if d𝑥 by d𝑡 represents the horizontal velocity of our function, then the rate of change in the horizontal velocity with respect to time must tell us the horizontal acceleration of our particle. And the same must be true in the vertical case. The rate of change in the vertical velocity will give us the vertical acceleration of our particle. So let’s use this to find the horizontal and vertical acceleration of our particle in terms of 𝑡.

We can find the horizontal acceleration of our particle by differentiating d𝑥 by d𝑡 with respect to 𝑡. That’s the derivative of 𝑡 squared minus six 𝑡 with respect to 𝑡. And we see this is a polynomial in terms of 𝑡. We can differentiate it using the power rule for differentiation. We multiply it by the exponent and then reduce the exponent by one. This gives us two 𝑡 minus six. We can do the same to find the vertical acceleration of our particle at the time 𝑡. We differentiate 𝑡 squared plus five 𝑡 minus three with respect to 𝑡. And this is also a polynomial. So we can differentiate this using the power rule for differentiation. This gives us two 𝑡 plus five.

Since we’ve now found the horizontal and vertical acceleration of our particle at the time 𝑡, we can represent the acceleration of our particle as a vector. We can represent the acceleration of our particle as two 𝑡 minus six 𝐢 plus two 𝑡 plus five 𝐣. However, the question is asking us about the acceleration, not the acceleration vector. And we can find the acceleration by taking the magnitude of our acceleration vector. And the magnitude of this vector is equal to the square root of the horizontal acceleration squared plus the vertical acceleration squared.

Since we want to find the time to the nearest tenth at which our acceleration is equal to 64, we’ll set the acceleration equal to 64. And that’s equal to the square root of the horizontal acceleration squared, that’s two 𝑡 minus six squared, plus the vertical acceleration squared. That’s two 𝑡 plus five squared. We can distribute the exponent over our parentheses either using the FOIL method or binomial expansion. Doing this for the first term gives us four 𝑡 squared minus 24𝑡 plus 36. And for the second term, we get four 𝑡 squared plus 20𝑡 plus 25.

So we’re now solving 64 is equal to the square root of four 𝑡 squared minus 24𝑡 plus 36 plus four 𝑡 squared plus 20𝑡 plus 25. We can then simplify this by grouping the like terms. This gives us that 64 is equal to the square root of eight 𝑡 squared minus four 𝑡 plus 61. We’re still not at the point which we can solve for the value of 𝑡. However, we can simplify our problem by squaring both sides of the equation. This gives us that 4096 is equal to eight 𝑡 squared minus four 𝑡 plus 61. We can then subtract 4096 from both sides of this equation to get a quadratic equation.

Simplifying this means we need to solve the quadratic equation eight 𝑡 squared minus four 𝑡 minus 4035 is equal to zero. One way of solving this could be by using the quadratic formula. This would give us the following expression for our two solutions of 𝑡, which if we calculate, we get one negative and one positive answer for 𝑡. However, 𝑡 represents the time in this case. So we only want the positive answer, which, to the nearest tenth, is 22.7.

Therefore, we’ve shown if a curve is moving on a path defined by the parametric equations d𝑥 by d𝑡 is equal to 𝑡 squared minus six 𝑡 and d𝑦 by d𝑡 is equal to 𝑡 squared plus five 𝑡 minus three. Then the time to the nearest tenth at which the acceleration of the particle will be 64 is 22.7.