Asked by norma
find the rate of change of the distance between the origin and a moving point on the graph of y=x^2+2 if ds/dt=5 centimeters per second.
Answers
Answered by
Reiny
I will assume you meant
dx/dt = 5, or else the answer to your question is given
s = √(x^2 + y^2)
= (x^2 + (x^2+2)^2)^.5
= (x^4 + 5x^2 + 4)^.5
ds/dt = (1/2)(x^4 + 5x^2 + 4)(4x^3 + 10x)(dx/dt)
so after you plug in the value, not much else can be done
Something fishy about the question, was there an x value given?
dx/dt = 5, or else the answer to your question is given
s = √(x^2 + y^2)
= (x^2 + (x^2+2)^2)^.5
= (x^4 + 5x^2 + 4)^.5
ds/dt = (1/2)(x^4 + 5x^2 + 4)(4x^3 + 10x)(dx/dt)
so after you plug in the value, not much else can be done
Something fishy about the question, was there an x value given?
Answered by
Reiny
forgot the exponent ... should say'
ds/dt = (1/2)(x^4 + 5x^2 + 4)^(-1/2)(4x^3 + 10x)(dx/dt)
ds/dt = (1/2)(x^4 + 5x^2 + 4)^(-1/2)(4x^3 + 10x)(dx/dt)
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