Differentiate both sides of the given equation with respect to time [tex]t[/tex]:
[tex]y=4x^2+1\implies\dfrac{\mathrm dy}{\mathrm dt}=8x\,\dfrac{\mathrm dx}{\mathrm dt}[/tex]
[tex]x[/tex] increases at a rate of 2 units/second, which means [tex]\dfrac{\mathrm dx}{\mathrm dt}=2[/tex]. So when [tex]x=1[/tex], we find
[tex]\dfrac{\mathrm dy}{\mathrm dt}=8\cdot2=16[/tex]
which is positive and thus increasing, so the answer is C.
The rate is at which y is changing when x = 1 is Option (C) Increasing 16 unit/sec.
Given curve is = [tex]y = 4x^{2} + 1[/tex]
Differentiating the above equation with respect to t where t is the time period.
[tex]\frac{dy}{dt} = 8x\frac{dx}{dt}[/tex]
Given x value is increasing at the rate of 2 units per second.
We have to find at what rate y is changing when x = 1.
This means [tex]\frac{dx}{dt} = 2 , x = 1[/tex]
⇒ [tex]\frac{dy}{dt} = 8*1*2[/tex]
∴ [tex]\frac{dy}{dt} = 16[/tex]
Therefore the rate is at which y is changing when x = 1 is Option (C) Increasing 16 unit/sec.
To learn more about rate change of a quantity, refer:
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