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2 months ago

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Assume that x and y are both differentiable functions of t and find the required values of dy/dt and dx/dt. x2 + y2 = 25 (a) Fin

d dy/dt, given x = 3, y = 4, and dx/dt = 5. dy/dt = -3/4 Incorrect: Your answer is incorrect. (b) Find dx/dt, given x = 4, y = 3, and dy/dt = –5.

1 answer:

Zina [12.3K]2 months ago

5 0

Answer:

(a) $\frac{dy}{dt}=-3\frac{3}{4}$

(b) $\frac{dx}{dt}=3\frac{3}{4}$

Step-by-step explanation:

$x^{2} +y^{2}=25$

Calculate $\frac{d}{dt}$ for each term.

$\frac{d}{dt}(x^{2})+\frac{d}{dt}(y^{2})=\frac{d}{dt}(25)\\\\(\frac{d}{dx}(x^{2})*\frac{dx}{dt}) +(\frac{d}{dy}(y^{2})*\frac{dy}{dt})=\frac{d}{dt}(25)\\\\2x\frac{dx}{dt} +2y\frac{dy}{dt} = 0\\\\$

For Question a

$2y\frac{dy}{dt}=-2x\frac{dx}{dt}\\\\\frac{dy}{dt}=\frac{-2x\frac{dx}{dt}}{2y} \\\\\frac{dy}{dt}=-\frac{x}{y}\frac{dx}{dt}$

With x = 3, y = 4, and dx/dt = 5.

$\frac{dy}{dt}=-\frac{3}{4}*5=-\frac{15}{4}\\ \\\frac{dy}{dt}=-3\frac{3}{4}$

For Question b

$2x\frac{dx}{dt}=-2y\frac{dy}{dt}\\\\\frac{dx}{dt}=\frac{-2y\frac{dy}{dt}}{2x} \\\\\frac{dx}{dt}=-\frac{y}{x}\frac{dy}{dt}$

Given x = 4, y = 3, and dx/dt = -5.

$\frac{dx}{dt}=-\frac{3}{4}*-5=\frac{15}{4}\\ \\\frac{dx}{dt}=3\frac{3}{4}$

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