A particle moves along the curve y=1+x3‾‾‾‾‾‾√. As it reaches the point (2,3), the y-coordinate is increasing at the rate of 4 cm/s. How fast is the x-coordinate of the point changing at this instant?

Answer:x-coordinate is changing by 2 cm/sStep-by-step explanation:Given equation of the curve,[tex]y=\sqrt{1+x^3}-----(1)[/tex]Differentiating with respect to t ( time ),[tex]\frac{dy}{dt} =\frac{3x^2}{2} \frac{1}{\sqrt{1+x^3}}\frac{dx}{dt}[/tex][tex]\frac{dy}{dt}=\frac{3x^2}{2y}\frac{dx}{dt}[/tex] ( From equation (1) ),We have given,[tex]\frac{dy}{dt}[/tex]=4 cm/s and x = 2, y = 3,[tex]4=\frac{3(2)^2}{2(3)}\frac{dx}{dt}[/tex][tex]4=\frac{12}{6}\frac{dx}{dt}[/tex][tex]4=2\frac{dx}{dt}[/tex][tex]\implies \frac{dx}{dt}=\frac{4}{2}=2\text{ cm per sec}[/tex]