Differential Equations — Quiz 4
Practice Differential Equations — Quiz 4 on The School of Mathematics (Differential Equations) with 5 scored questions. This set covers problems such as: “Which of the following differential equations is not logistic?”; “The slope field for F'(x)=e^{-x^2} is shown above with the particular solution F(0)=0 superimposed. With a …”; “According to Newton's law of cooling, the temperature of an object decreases at a rate proportional to the …”. Work carefully, check explanations for misses, then retake to track improvement before test day.
Questions in this quiz (5)
Which of the following differential equations is not logistic?
- $$\frac{dP}{dt}=0.02P(500-P)$$
- $$P'=P-0.001P^2$$
- $$\frac{dx}{dt}=0.5x(200-x)$$
- $$\frac{dN}{dt}=0.04(300-N)$$
The slope field for $$F'(x)=e^{-x^2}$$ is shown above with the particular solution $$F(0)=0$$ superimposed. With a graphing calculator,$$\lim_{x\to\infty}F(x)$$ to $$3$$ decimal places is:
- $$0.886$$
- $$0.987$$
- $$1.000$$
- $$\infty$$
According to Newton's law of cooling, the temperature of an object decreases at a rate proportional to the difference between its temperature and that of the surrounding air. Suppose a cup of tea at a temperature of $$90^\circ C$$ is placed in a room where the temperature is $$20^\circ C$$. The differential equation is:$$\frac{dT}{dt}=-k(T-20)$$where $$T$$ is the temperature of the tea in $$^\circ C$$ and $$t$$ is time in minutes. If the tea cools to $$70^\circ C$$ in $$10$$ minutes, then its temperature is given by the equation:
- $$T=20+70e^{-0.034t}$$
- $$T=20+70e^{-0.046t}$$
- $$T=20+90e^{-0.046t}$$
- $$T=90e^{-0.046t}$$
A bowl of soup at temperature $$190^\circ F$$ is placed on a table in a room at $$70^\circ F$$. The differential equation for its temperature at time $$t$$ in minutes is$$\frac{dy}{dt}=-0.09(y-70);$$ $$y(0)=190$$After $$10$$ minutes, the temperature in $$^\circ F$$ of the soup is approximately:
- $$112$$
- $$119$$
- $$125$$
- $$140$$
$$[-4,4]$$ x $$[-4,4]$$A solution curve has been superimposed on the slope field shown above. The solution is for the differential equation and initial condition:
- $$\frac{dy}{dx}=\tan x,$$ $$y(0)=0$$
- $$\frac{dy}{dx}=1+x^2,$$ $$y(0)=0$$
- $$\frac{dy}{dx}=\frac{1}{1+x^2},$$ $$y\left(\frac{\pi}{4}\right)=1$$
- $$\frac{dy}{dx}=1+y^2,$$ $$y(0)=0$$