A moving particle has a height given by y = 14(x-3)^2 + 3x + 9. What is the maximum height of this particle.

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justaguide | College Teacher | (Level 2) Distinguished Educator

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The height of the particle (expressed by y) is given in terms of its horizontal displacement x by `y = 14(x-3)^2 + 3x + 9` .

To determine the maximum height of the particle, find the derivative of y with respect to x.

For `y = 14(x-3)^2 + 3x + 9` ,

`dy/dx = 14*2*(x - 3) + 3`

But `(d^2y)/(dx^2) = 28` which is positive for all x. The graph of `y = 14(x-3)^2 + 3x + 9` has an extreme value which is a minimum. The height of the particle can increase to any extent.

If the lowest height of the particle has to be determined, solve `dy/dx` = 0 for x.

`14*2*(x - 3) + 3 = 0`

=> `14*2*(x - 3) = -3 `

=> `x - 3 = -3/28`

=> `x = 3 - 3/28`

=> `x = 81/28`

At x = `81/28` , y = `14*(81/28-3)^2 + 3*81/28 + 9`

= `999/56`

The lowest height of the particle is `999/56` , there is no limit to how high it can rise.

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tonys538 | Student, Undergraduate | (Level 1) Valedictorian

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The vertical height of the moving particle y, as a function of the horizontal distance traveled x is given by the function y = 14(x-3)^2 + 3x + 9.

Now, y = 14(x-3)^2 + 3x + 9

= 14*(x^2 - 6x + 9) + 3x + 9

= 14x^2 - 81x + 135

This is the equation of a parabola. It can be written in the vertex form as y = a*(x - h)^2 + k where (h, k) is the vertex.

y = 14x^2 - 81x + 135

= 14(x^2 - (81/14)x) + 135

= 14(x^2 - 2*x*(81/28) + (81/28)^2) + 135 - 14*(81/28)^2

= 14(x^2 - (81/28))^2 + 999/56

Now 14 is a positive integer. The parabola defined by y = 14(x-3)^2 + 3x + 9 is one that opens upwards.

As a result, the maximum height of the particle is not limited. As x increases beyond 81/28, the value of y increases and continues to do so forever.

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