Question:
The rate at which a baby bird gains weight is proportional to the difference between its adult weight and its current weight. At time t = 0, when the bird is first weighed, its weight is 20 grams. If B (t) is the weight of the bird, in grams, at time t days after it is first weighed, then
\(\frac{dB}{dt}= \frac{1}{5}(100-B).\)
Let y = B (t) be the solution to the differential equation above with initial condition B(0) = 20.
(a) Is the bird gaining weight faster when it weighs 40 grams or when it weighs 70 grams? Explain your reasoning.
(b) Find \(\frac{d^{2}B}{dt^{2}}\) in terms of B. Use \(\frac{d^{2}B}{dt^{2}}\) to explain why the graph of B cannot resemble the following graph.
(c) Use separation of variables to find y = B(t), the particular solution to the differential equation with initial condition B(0) = 20.
▶️Answer/Explanation
Ans:
(a) When it is 40 grams : \(\frac{dB}{dt}=\frac{1}{5}(100-40)=12\) g/day
when it is 70 grams: \(\frac{dB}{dt}=\frac{1}{5}(100-70)=6\) g/day
So the bird is gaining weight faster when it weighs 40 grams.
(b)
\(\frac{dB}{dt}=20-\frac{1}{5}B\)
\(\frac{d^{2}B}{dt^{2}}=-\frac{1}{5}.\frac{dB}{dt}\)
\(=-\frac{1}{5}(20-\frac{1}{5}B)\)
\(=\frac{1}{25}B-4\)
\(\frac{1}{25}B-4>0\)
B > 100
So, the graph cannot be concave up when weight is below 100 g.
(c)
\(\frac{dB}{dt}=\frac{1}{5}(100-B)\)
\(\frac{1}{\frac{1}{5}(100-B)}dB = dt\)
\(\frac{1}{100-B}dB = dt\)
\(\int \frac{1}{100-B}dB = \int 1dt\)
-5 In [100-B] = t + c
\(In (100-B) = – \frac{1}{5}(t+c)\)
\(100-B = e^{- \frac{1}{5}(t+c)}\)
\(B = 100-e^{- \frac{1}{5}(t+c)}\)
\(20 = 100-e^{- \frac{1}{5}c}\)
\(e^{- \frac{1}{5}c}=80\)
\(- \frac{1}{5}c= ln 80\)
\(c = -5 ln 80\)
\(\partial B= 100-e^{- \frac{1}{5}c+-5 ln 80}\)
Question:
A cylindrical barrel with a diameter of 2 feet contains collected rainwater, as shown in the figure above. The water drains out through a valve (not shown) at the bottom of the barrel. The rate of change of the height h of the water in the barrel with respect to time t is modeled by \(\frac{dh}{dt}=-\frac{1}{10}\sqrt{h},\) where h is measured in feet and t is measured in seconds. (The volume V of a cylinder with radius r and height h is V = πr2h.)
(a) Find the rate of change of the volume of water in the barrel with respect to time when the height of the water is 4 feet. Indicate units of measure.
(b) When the height of the water is 3 feet, is the rate of change of the height of the water with respect to time increasing or decreasing? Explain your reasoning.
(c) At time t = 0 seconds, the height of the water is 5 feet. Use separation of variables to find an expression for h in terms of t.
▶️Answer/Explanation
Ans:
(a)
h = 4
\(\frac{dh}{dt}=\frac{-1}{10}\sqrt{4}=\frac{-1}{5}\) \(\frac{dv}{dt}=\pi \left [ r^{2}\frac{dh}{dt}+h(2r)\left ( \frac{dr}{dt} \right ) \right ]\)
\(\frac{dv}{dt}=?\) \(\frac{dv}{dt}=\pi \left [ (1)\left ( \frac{-1}{5} \right )+(4)(2)(0) \right ]\)
r = 1
\(\frac{dv}{dt}=0\) \(\frac{dv}{dt}=\frac{-\pi }{5}ft^{3}/sec\)
(b)
h = 3
\(\frac{d^{2}h}{dt^{2}}=?\) \(\frac{d^{2}h}{dt^{2}}=\frac{-1}{20}h^{-1/2}\frac{dh}{dt}=\frac{-1}{20\sqrt{3}}\left ( \frac{-1}{10}\sqrt{3} \right )=\frac{1}{200}\)
\(\frac{-1}{10}h^{1/2}\) The rate of change of height is increasing since \(\frac{d^{2}h}{dt^{2}}\) at h = 3 is positive.
1/20
(c)
\(\int h^{-1/2}dh=\int \frac{-1}{10}dt\)
\(\int h^{1/2}=\frac{-1}{10}t+c_{1}\)
\(h^{1/2}=\frac{-1}{20}t+c_{2}\)
\(h=\left ( \frac{-1}{20}t+c \right )^{2}\)
\(5=\left ( \frac{-1}{20}(0)+c \right )^{2}\)
\(c = \sqrt{5}\)
\(h = \left ( \frac{-1}{20}t+\sqrt{5} \right )^{2}\)
Question
Consider the differential equation \(\frac{dy}{dx}=\frac{x}{y} where y\neq 0\)
(a) The slope field for the given differential equation is shown below. Sketch the solution curve that passes through the point ( 3,- 1 ), and sketch the solution curve that passes through the point ( 1, 2) .(Note: The points ( 3, -1) and ( 1, 2) are indicated in the figure.)
(b) Write an equation for the line tangent to the solution curve that passes through the point ( 1, 2) .
(c) Find the particular solution y = f ( x) to the differential equation with the initial condition f ( 3 ), = -1 and state its domain
▶️Answer/Explanation
.
(a)\(g(1)=\int_{1}^{1}f(t)dt=0 \)and \(g'(1)=f(1)=0 \)(b) Since g'(x) = f(x), g is increasing on the interval 1≤ x ≤ because \(f(x) > 0 for 1 < x < \pi.\) (c) For \(0 < x < 2\pi, g'(x) = f(x) = 0 \)when x = 1, \pi. g’ f changes from negative to positive only at x = 1. The absolute minimum must occur at x = 1 or at the right endpoint. Since g(1) = 0 and \(g(2\pi)=\int_{1}^{2\pi}f(t)dt=\int_{1}^{\pi}f(t)dt+\int_{\pi}^{2\pi}f(t)<0\) by comparison of the two areas, the absolute minimum occurs at \(x = 2π. \)(d) Yes, the graph of g is tangent to the x-axis at x = 1 since g(1) = 0 and g'(1) = 0.