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A bug is moving along the right side of the parabola \( y=x^{2} \) at a rate such that its distance from the origin is increasing at \( 3 \mathrm{~cm} / \mathrm{min} \). a. At what rate is the \( x \)-coordinate of the bug increasing when the bug is at the point \( (6,36) \) ? b. Use the equation \( y=x^{2} \) to find an equation relating \( \frac{d y}{d t} \) to \( \frac{d x}{d t} \). c. At what rate is the \( y \)-coordinate of the bug increasing when the bug is at the point \( (6,36) \) ? a. At what rate is the \( x \)-coordinate of the bug increasing when the bug is at the point \( (6,36) \) ? The \( x \)-coordinate of the bug is increasing at a rate of (Type an exact answer, using radicals as needed.)

Ask by Erickson George. in the United States
Mar 12,2025

Solución de inteligencia artificial de Upstudy

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- a. The \( x \)-coordinate is increasing at \( \frac{3\sqrt{37}}{73} \, \text{cm/min} \). - b. \( \frac{dy}{dt} = 2x \frac{dx}{dt} \). - c. The \( y \)-coordinate is increasing at \( \frac{36\sqrt{37}}{73} \, \text{cm/min} \).

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The Deep Dive

To solve this, we can use the relationship between \( x \), \( y \), and the distance from the origin. The distance from the origin to the point \( (x, y) \) is given by \( d = \sqrt{x^2 + y^2} \). Given that \( y = x^2 \), we can express \( d \) specifically as \( d = \sqrt{x^2 + (x^2)^2} = \sqrt{x^2 + x^4} \). Now, differentiating \( d \) with respect to \( t \): \[ \frac{d}{dt}(d) = \frac{d}{dt}(\sqrt{x^2 + x^4}) = \frac{1}{2}(x^2 + x^4)^{-1/2}(2x\frac{dx}{dt} + 4x^3\frac{dx}{dt}) \] \[ \frac{dd}{dt} = \frac{(2x + 4x^3)\frac{dx}{dt}}{2\sqrt{x^2 + x^4}} \] From the problem, we know \( \frac{dd}{dt} = 3 \) cm/min when \( x = 6 \) and \( y = 36 \). Substituting those values into the equation gives: \[ 3 = \frac{(2(6) + 4(6)^3)\frac{dx}{dt}}{2\sqrt{6^2 + 36^2}} \] Calculating \( \sqrt{6^2 + 36^2} = \sqrt{36 + 1296} = \sqrt{1332} = 6\sqrt{37} \). Next, calculate: \[ 2(6) + 4(216) = 12 + 864 = 876 \] So we have: \[ 3 = \frac{876\frac{dx}{dt}}{12\sqrt{37}} \implies \frac{dx}{dt} = \frac{3 \cdot 12\sqrt{37}}{876} = \frac{36\sqrt{37}}{876} \] Simplifying gives \( \frac{dx}{dt} = \frac{1\sqrt{37}}{24.5} \), which is approximately \( 0.07416\) cm/min in radicals. Now, pertaining to the rate of increase of \( \frac{dy}{dt} \): Given \( y = x^2 \), differentiating gives \( \frac{dy}{dt} = 2x\frac{dx}{dt} \). Now substituting \( x = 6 \) and \( \frac{dx}{dt} = \frac{36\sqrt{37}}{876}\): \[ \frac{dy}{dt} = 2(6)\left(\frac{36\sqrt{37}}{876}\right) = \frac{432\sqrt{37}}{876} \approx 0.04433 \text{ cm/min in radicals.} \] Lastly, we'd find that as the bug moves along the parabola, understanding these rates helps visualize its motion in a dynamic way—like watching a tiny racecar zoom along a curvy track!

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