Section II Part A — Full Practice Exam 2
Practice Section II Part A — Full Practice Exam 2 on The School of Mathematics (Full Practice Exam 2) with 2 scored questions in about 30 min. This set covers problems such as: “The temperature on a winter day in Northfield is modeled by: T(H)=C+D\sin\left(\frac{\pi H}{24}\right) wher…”; “A population of algae grows in a pond. The rate of growth of the algae is proportional to the amount presen…”. Work carefully, check explanations for misses, then retake to track improvement before test day.
Questions in this quiz (2)
The temperature on a winter day in Northfield is modeled by:$$T(H)=C+D\sin\left(\frac{\pi H}{24}\right)$$where $$T$$ is the temperature in degrees Fahrenheit and $$H$$ is the number of hours from midnight, $$0\le H\le24$$.a. The temperature at midnight is $$10$$ and the temperature at noon is $$30$$. Find $$C$$ and $$D$$.b. Find the average temperature during the first $$8$$ hours of the day.c. Use the Trapezoidal Rule with $$4$$ equal subintervals to estimate$$\int_{0}^{8}T(H)dH$$.d. Find an expression for the rate at which the temperature is changing with respect to $$H$$.
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A population of algae grows in a pond. The rate of growth of the algae is proportional to the amount present at time $$t$$, so that $$\frac{dA}{dt}=kA$$ where $$k$$ is a positive constant and $$A(t)$$ is measured in tons.a. Find an expression for $$A(t)$$ in terms of $$t$$, given that the initial amount of algae is $$50$$ tons and that after $$2$$ years the amount has increased to $$80$$ tons.b. After how many years will the amount of algae reach $$200$$ tons?c. Suppose that algae-eating organisms are introduced into the pond and remove algae at a constant rate of $$30$$ tons per year. Write a differential equation that models the new situation, and determine how long it will take for the pond to become completely free of algae.
- a. $$A(t)=50\left(\frac{8}{5}\right)^{t/2}$$ b. $$t=\frac{2\ln4}{\ln\frac{8}{5}}$$ c. $$\frac{dA}{dt}=kA-30$$ and $$t=\frac{1}{k}\ln\left(\frac{30}{30-50k}\right)$$ with $$k=\frac{1}{2}\ln\frac{8}{5}$$
- a. $$A(t)=50\left(\frac{8}{5}\right)^{t/2}$$ b. $$t=\frac{2\ln4}{\ln\frac{8}{5}}$$ c. $$\frac{dA}{dt}=kA-30$$ and $$t=\frac{1}{k}\ln\left(\frac{30}{30-50k}\right)$$ with $$k=\frac{1}{2}\ln\frac{8}{5}$$