Respuesta :
Answer:
x-coordinate is changing by 2 cm/s
Step-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]
The x-coordinate of the point is changing at 2cm/s this instant
Given the function;
[tex]y=\sqrt{1+x^3}[/tex]
The rate of change of the function is expressed as:
[tex]\frac{dy}{dt} =\frac{dy}{dx} \times \frac{dx}{dt}[/tex]
Get the differential of y with respect to x:
[tex]y = (1+x^3)^{\frac{1}{2} }\\\frac{dy}{dx} = \frac{1}{2} (1+x^3) ^{-1/2} \times 3x^2\\\frac{dy}{dx} =\frac{3}{2}x^2(1+x^3)^{-1/2}\\\frac{dy}{dt}=\frac{3x^2}{2y}[/tex]
Substitute into the rate of change to have:
[tex]\frac{dy}{dt} =\frac{3x^2}{2y}\frac{dx}{dt}[/tex]
Given the following parameters:
[tex]\frac{dy}{dt} = 4cm/s\\x=2\\y=3[/tex]
Substitute into the rate of change formula to get dx/dt
[tex]\frac{dy}{dt} =\frac{3(2)^2}{2(3)} \times \frac{dx}{dt} \\4 = \frac{12}{6} \times\frac{dx}{dt} \\24 = 12\frac{dx}{dt}\\\frac{dx}{dt} = \frac{24}{12}\\\frac{dx}{dt} = 2cm/s[/tex]
Hence the x-coordinate of the point is changing at 2cm/s this instant
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