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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 kirsten
A particle moves along the curve below.
y =
sqrt(8 + x^3)
As it reaches the point (2, 4), the y-coordinate is increasing at a rate of 5 cm/s. How fast is the x-coordinate of the point changing at that instant?
y =
sqrt(8 + x^3)
As it reaches the point (2, 4), the y-coordinate is increasing at a rate of 5 cm/s. How fast is the x-coordinate of the point changing at that instant?
Answers
Answered by
Steve
y = √(8+x^3)
dy/dt = (3x^2 / 2√(8+x^3)) dx/dt
Now just plug in dy/dt=5 and x=2 to get dx/dt
dy/dt = (3x^2 / 2√(8+x^3)) dx/dt
Now just plug in dy/dt=5 and x=2 to get dx/dt
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