If f and g are functions of time, and at time t = 3, f equals 8 and is rising at a rate of 3 units per second, and g equals 5 and is rising at a rate of 8 units per second, then the product fg equals __ and is rising at a rate of __ units per second.
I got 40 for the first answer, but need help with the second please!
To find the rate at which the product fg is rising, we can use the product rule of differentiation.
The product rule states that if we have two functions u(t) and v(t) that depend on time t, then the derivative of their product is given by:
d/dt (u(t)*v(t)) = u'(t)*v(t) + u(t)*v'(t)
In this case, let u(t) = f(t) and v(t) = g(t). We are given that at time t = 3, f equals 8 and is rising at a rate of 3 units per second, and g equals 5 and is rising at a rate of 8 units per second.
Therefore, we have:
u(3) = f(3) = 8
u'(3) = rate of change of f at t = 3 = 3 units per second
v(3) = g(3) = 5
v'(3) = rate of change of g at t = 3 = 8 units per second
Now, we can substitute these values into the product rule formula:
d/dt (u(t)*v(t)) = u'(t)*v(t) + u(t)*v'(t)
d/dt (fg) = u'(t)*v(t) + u(t)*v'(t)
At t = 3, we have:
d/dt (fg) = u'(3)*v(3) + u(3)*v'(3)
d/dt (fg) = 3*5 + 8*8
d/dt (fg) = 15 + 64
d/dt (fg) = 79
So, the rate at which the product fg is rising is 79 units per second.
Therefore, the answer is:
The product fg equals 40 and is rising at a rate of 79 units per second.
To find the rate at which the product fg is rising, we can use the product rule of differentiation. The product rule states that if u(x) and v(x) are functions of x, then the derivative of their product is given by:
(d/dx)(u(x) * v(x)) = u'(x) * v(x) + u(x) * v'(x)
In this case, f and g are functions of time, so we need to take into account their rates of change with respect to time, denoted as f'(t) and g'(t), respectively.
Since f is rising at a rate of 3 units per second, we have f'(t) = 3.
Similarly, g is rising at a rate of 8 units per second, so g'(t) = 8.
Now, let's find the rate at which the product fg is rising by applying the product rule:
(d/dt)(f * g) = f'(t) * g(t) + f(t) * g'(t)
Substituting the given values:
(d/dt)(f * g) = 3 * g(t) + f(t) * 8
At time t = 3, f equals 8 and g equals 5, so we can substitute these values into the equation:
(d/dt)(f * g) = 3 * 5 + 8 * 8 = 15 + 64 = 79
Therefore, the product fg is rising at a rate of 79 units per second.
I'm sure your text includes the derivation of the formula that the rate of change of fg is
∆(fg)/∆t = f * ∆g/∆t + g * ∆f/∆t
In this case, the rate of change of fg is thus
8*5 + 5*3 = 55 units/sec