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A curve has equation y=160(3x+1)2y=\frac{1}{60}(3x+1)^{2} and a point is moving along the curve. Find the xx-coordinate of the point on the curve at which the xx - and yy-coordinates are increasing at the same rate. \quad

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Q. A curve has equation y=160(3x+1)2y=\frac{1}{60}(3x+1)^{2} and a point is moving along the curve. Find the xx-coordinate of the point on the curve at which the xx - and yy-coordinates are increasing at the same rate. \quad
  1. Find Derivative of y: To find where the xx and yy coordinates are increasing at the same rate, we need to find the derivative of yy with respect to xx, which will give us the rate of change of yy.dydx=(160)2(3x+1)(3)\frac{dy}{dx} = \left(\frac{1}{60}\right) \cdot 2 \cdot (3x + 1) \cdot (3)
  2. Simplify Derivative: Simplify the derivative.\newlinedydx=130×(3x+1)×3\frac{dy}{dx} = \frac{1}{30} \times (3x + 1) \times 3\newlinedydx=110×(3x+1)\frac{dy}{dx} = \frac{1}{10} \times (3x + 1)
  3. Set Derivative Equal: Set the derivative equal to 11 since the rate of change of yy must equal the rate of change of xx, which is 11.110(3x+1)=1\frac{1}{10} \cdot (3x + 1) = 1
  4. Solve for x: Solve for x.\newline3x+1=103x + 1 = 10\newline3x=93x = 9\newlinex=3x = 3

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