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A particle moves along the curve y=square sqrt 1+x^3. As it reaches the point (2,3), the y-coordinate is increasing ata a rate...Asked by Tay
a particle moves along the curve y= sqrt 1+x cubed. As it reaches the point (2,3) the y-corrdinate is increasing at a rate of 4cm/s. How fast is the x-coordinate of the point changing at that instant?
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Answered by
bobpursley
y= sqrt(1+x^3)
dy/dt= 1/2 * 1/sqrt(1+x^3)* 3x^2 dx/dt
dy/dt= 3x^2/2y * dx/dt
dx/dt= 2y dy/dt * 1/3x^2
you are given x, y, dy/dt, solve fod dx/dt.
check my work, I did it in a hurry
dy/dt= 1/2 * 1/sqrt(1+x^3)* 3x^2 dx/dt
dy/dt= 3x^2/2y * dx/dt
dx/dt= 2y dy/dt * 1/3x^2
you are given x, y, dy/dt, solve fod dx/dt.
check my work, I did it in a hurry
Answered by
Tay
hey thanks we are checking it now:)
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