Suppose f(t) is the weight (in grams) of a solid sitting in a beaker of water and the solid is dissolving at a rate (in grams/minute) of f'(t)=-4 f(t)(3 + f(t) ).
If there is 4 grams of solid at time t=2 minutes, use a linear approximation to estimate the solid's weight 1 second later.
To estimate the solid's weight 1 second later using linear approximation, we can use the tangent line to the curve at the given point (t=2, f(2)=4).
First, we need to find the equation of the tangent line. To do this, we'll find the value of the first derivative f'(t) at t=2 and then use the point-slope formula for a line.
Given f'(t) = -4f(t)(3 + f(t)), we can substitute t=2 and f(t)=4 into the equation to find the slope of the tangent line.
f'(2) = -4(4)(3 + 4) = -4(4)(7) = -4(28) = -112
The slope of the tangent line is -112.
Next, we'll use the point-slope formula for a line with the point (2, 4) and the slope -112:
y - y1 = m(x - x1)
where (x1, y1) = (2, 4) and m = -112.
y - 4 = -112(x - 2)
Simplifying this equation, we get:
y - 4 = -112x + 224
Adding 4 to both sides:
y = -112x + 228
Now, we can use this equation to estimate the solid's weight 1 second (t = 2 + 1/60 minutes) later.
To do this, we substitute t = 2 + 1/60 into the equation:
f(t) = -112(2 + 1/60) + 228
Calculating this expression gives an estimate of the solid's weight 1 second later.
To use the linear approximation, we first need to find the derivative of f(t) with respect to t, since we are given the derivative f'(t)=-4 f(t)(3 + f(t)).
Let's find the value of f(t) at t=2 minutes:
f(2) = 4 grams
Now, we can find the derivative of f(t) at t=2 minutes using the given derivative:
f'(2) = -4 f(2)(3 + f(2))
Substituting the value of f(2):
f'(2) = -4 * 4 * (3 + 4)
= -4 * 4 * 7
= -112 grams/minute
To estimate the weight of the solid 1 second later, we can use the linear approximation:
f(2+1/60) ≈ f(2) + f'(2) * (1/60)
Substituting the values:
f(2+1/60) ≈ 4 + (-112) * (1/60)
Now, let's calculate the estimation:
f(2+1/60) ≈ 4 - 112/60
≈ 4 - 56/30
≈ 4 - 28/15
≈ 60/15 - 28/15
≈ 32/15
Therefore, the linear approximation estimates the solid's weight to be approximately 32/15 grams 1 second later.