suppose that the amount in grams of a radioactive substance present at time t (in years) is given by A (t) = 160e-.70t. Find the rate of decay of the quantity present at the time when t = 4
To find the rate of decay at a specific time, we need to determine the derivative of the given function A(t).
Given: A(t) = 160e^(-0.70t)
To find the derivative, let's use the chain rule. The derivative of A(t) with respect to t can be expressed as follows:
dA/dt = (d/dt)(160e^(-0.70t))
To differentiate this, we use the power rule and the chain rule:
dA/dt = -0.70 * 160e^(-0.70t)
Now, we need to find the rate of decay when t = 4:
dA/dt = -0.70 * 160e^(-0.70(4))
Simplifying this expression:
dA/dt = -0.70 * 160e^(-2.8)
Using a calculator, we can approximate the value of e^(-2.8):
dA/dt ≈ -0.70 * 160 * 0.05947
Calculating further:
dA/dt ≈ -6.29144
Hence, the rate of decay of the quantity present at t = 4 is approximately -6.29144 grams per year.
To find the rate of decay of the quantity present at a specific time, we need to find the derivative of the function that describes the quantity as a function of time.
In this case, the function that describes the quantity of the radioactive substance is A(t) = 160e^(-0.70t).
To find the derivative, we'll use the chain rule. The chain rule states that if we have a composite function, f(g(x)), then the derivative of f(g(x)) with respect to x is given by f'(g(x)) * g'(x).
Let's find the derivative step by step:
1. Start with A(t) = 160e^(-0.70t).
2. Take the derivative of the outer function, which is e^(-0.70t). The derivative of e^x with respect to x is e^x, so the derivative of e^(-0.70t) with respect to t is -0.70 * e^(-0.70t).
3. Multiply the derivative of the outer function by the derivative of the inner function, which is -0.70, since we're differentiating with respect to t: -0.70 * -0.70 * e^(-0.70t).
4. Simplify the expression: 0.49 * e^(-0.70t).
Now we have the derivative of A(t) with respect to t. To find the rate of decay at a specific time when t = 4, we substitute t = 4 into the derivative.
Plug t = 4 into the derivative expression:
0.49 * e^(-0.70t) = 0.49 * e^(-0.70 * 4) ≈ 0.49 * e^(-2.8) ≈ 0.49 * 0.0608101 ≈ 0.029820 grams per year.
Therefore, the rate of decay at t = 4 is approximately 0.029820 grams per year.
A ' (t) = -.7(160) e^(-.7t)
so A ' (4) = -.7(160) e^-2.8
= -6.81
insert the units