Video: Finding the Point on the Curve of a Quadratic Function That Satisfies a Relation between the Rate of Change of Its Coordinates

A particle moves along the curve 𝑦 = 3π‘₯Β² βˆ’ 2π‘₯ βˆ’ 6. At what point is the rate of change in its 𝑦-coordinate four times the rate of change in its π‘₯-coordinate?

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Video Transcript

A particle moves along the curve 𝑦 equals three π‘₯ squared minus two π‘₯ minus six. At what point is the rate of change in its 𝑦-coordinate four times the rate of change in its π‘₯-coordinate?

The rate of change of the 𝑦-coordinate is equal to d𝑦 dπ‘₯. This means we need to differentiate our curve. The general rule for differentiating states that if 𝑦 is equal to π‘Ž multiplied by π‘₯ to the power of 𝑛, then d𝑦 by dπ‘₯ is equal to 𝑛 multiplied by π‘Ž multiplied by π‘₯ to the power of 𝑛 minus one. We multiply the power by the coefficient and reduce the power by one.

In this question, 𝑦 is equal to three π‘₯ squared minus two π‘₯ minus six. Differentiating the first term gives us six π‘₯ to the power of one as two multiplied by three is equal to six and reducing a power by one gives us one. Six π‘₯ to the power of one is the same as six π‘₯. Differentiating the second term gives us negative two π‘₯ to the power of zero as one multiplied by negative two is negative two, and reducing a power by one gives us zero. π‘₯ to the power of zero is equal to one. Therefore, negative two π‘₯ differentiates to negative two.

Any constant term differentiated gives us zero. Therefore, the differential of negative six is zero. If 𝑦 is equal to three π‘₯ squared minus two π‘₯ minus six, then d𝑦 by dπ‘₯ is equal to six π‘₯ minus two. Six π‘₯ minus two is the rate of change of 𝑦. We’re told that this is four times the rate of change in the π‘₯-coordinate. This can be written as an equation, six π‘₯ minus two is equal to four multiplied by π‘₯.

Adding two to both sides of this equation gives us six π‘₯ is equal to four π‘₯ plus two. We can then subtract four π‘₯ from both sides. This gives us two π‘₯ is equal to two. Finally, dividing both sides of this equation by two gives us π‘₯ is equal to one. We have worked out the π‘₯-coordinate and now need to work out the 𝑦-coordinate.

We do this by substituting π‘₯ equals one into our initial equation. This gives us 𝑦 is equal to three multiplied by one squared minus two multiplied by one minus six. One squared is equal to one. And multiplying this by three gives us three. Negative two multiplied by one is equal to negative two. This means that 𝑦 is equal to three minus two minus six. This gives us a 𝑦-coordinate of negative five. The point, or coordinate, at which the rate of change in the 𝑦-coordinate is four times the rate of change in the π‘₯-coordinate is one, negative five.

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