Question

1. the displacement of a particle at a time t, denoted by g(t), is defined implicitly as a function of t. in particular, the particle is moving in the plane along the cardioid shaped path given to the right; this path is described by the implicit equation, 9((t - 1)^2 + g^2) = ((t - 1)^2 + g^2 - 3(t - 1))^2 where g(t) is measured in nanometers per second. use implicit differentiation to determine the velocity of the particle when it is at the point (1, 3).

Explanation:

Step1: Let \(x = t - 1\), \(y = g(t)\)

The equation becomes \(9(x^{2}+y^{2})=(x^{2}+y^{2}-3x)^{2}\)

Step2: Differentiate both sides with respect to \(t\)

Using the chain - rule.
Left - hand side:

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Since \(\frac{dx}{dt}=1\), it is \(18(x + y\frac{dy}{dt})\)
Right - hand side:
Let \(u=x^{2}+y^{2}-3x\), then the right - hand side is \(u^{2}\)
\(\frac{d}{dt}(u^{2}) = 2u\frac{du}{dt}\)
\(\frac{du}{dt}=(2x - 3)\frac{dx}{dt}+2y\frac{dy}{dt}=(2x - 3)+2y\frac{dy}{dt}\)
So \(\frac{d}{dt}(u^{2})=2(x^{2}+y^{2}-3x)[(2x - 3)+2y\frac{dy}{dt}]\)

Step3: Substitute \(x = 1 - 1=0\), \(y = 3\) into the differentiated equation

Left - hand side: \(18(0 + 3\frac{dy}{dt})=54\frac{dy}{dt}\)
Right - hand side:
\(u=x^{2}+y^{2}-3x=0 + 9-0 = 9\)
\(2u[(2x - 3)+2y\frac{dy}{dt}]=2\times9[(- 3)+6\frac{dy}{dt}]=18(-3 + 6\frac{dy}{dt})\)
So \(54\frac{dy}{dt}=18(-3 + 6\frac{dy}{dt})\)

Step4: Solve for \(\frac{dy}{dt}\)

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Answer:

The velocity of the particle at the point \((1,3)\) is \(1\) nanometer per second.