Using implicit differentiation, it is found that:
[tex]\frac{dx}{dt} = 4 + 3\sqrt{3}[/tex]
The equation of the circle is:
[tex]x^2 + y^2 = 16[/tex]
We are given x = 2, then:
[tex]2^2 + y^2 = 16[/tex]
[tex]y^2 = 12[/tex]
[tex]y = \pm \sqrt{12}[/tex]
First quadrant, thus, positive:
[tex]y = \sqrt{12}[/tex]
[tex]y = 2\sqrt{3}[/tex]
Applying implicit differentiation, we have that:
[tex]2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 16[/tex]
We are given that [tex]frac{dy}{dt} = -3[/tex], then:
[tex]2(2)\frac{dx}{dt} - 12\sqrt{3} = 16[/tex]
[tex]4\frac{dx}{dt} = 16 + 12\sqrt{3}[/tex]
[tex]\frac{dx}{dt} = \frac{16 + 12\sqrt{3}}{4}[/tex]
[tex]\frac{dx}{dt} = 4 + 3\sqrt{3}[/tex]
A similar problem is given at https://brainly.com/question/9543179
