3x - 2y + 3z = 6
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You are to find the rate of change of B with respect to t, dB/dt. Differentiate the first equation with respect to t.
dB/dt = 0.007(2/3)(W^-1/3)(dW/dt)
We have to determine W and dW/dt.
For W when L = 18 cm, we use the second equation
W = 0.12(18^2.53) = 1,800 g
For dW/dt, differentiate the second equation with respect to t.
dW/dt = 0.12(2.53)(L^1.53)(dL/dt)
We still have to find dL/dt from the given information. The solution would be:
dL/dt = (20 cm - 15 cm)/10,000,000 years = 5×10⁻⁷ cm/y
So,
dW/dt = 0.12(2.53)(18^1.53)(5×10⁻⁷) = 1.26×10⁻⁵ g/y
Now that we know W and dW/dt, we can finally solve for dB/dt.
dB/dt = 0.007(2/3)(W^-1/3)(dW/dt)
dB/dt = 0.007(2/3)(1800^-1/3)(1.26×10⁻⁵) = <em>1.04×10⁻⁸ g/y</em>
dB/dt = 0.007(2/3)(W^-1/3)(dW/dt)
We have to determine W and dW/dt.
For W when L = 18 cm, we use the second equation
W = 0.12(18^2.53) = 1,800 g
For dW/dt, differentiate the second equation with respect to t.
dW/dt = 0.12(2.53)(L^1.53)(dL/dt)
We still have to find dL/dt from the given information. The solution would be:
dL/dt = (20 cm - 15 cm)/10,000,000 years = 5×10⁻⁷ cm/y
So,
dW/dt = 0.12(2.53)(18^1.53)(5×10⁻⁷) = 1.26×10⁻⁵ g/y
Now that we know W and dW/dt, we can finally solve for dB/dt.
dB/dt = 0.007(2/3)(W^-1/3)(dW/dt)
dB/dt = 0.007(2/3)(1800^-1/3)(1.26×10⁻⁵) = <em>1.04×10⁻⁸ g/y</em>