7:[["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"Quiz\",\"name\":\"Quiz 3\",\"description\":\"Practice problems on differential equations\",\"url\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-differential-equations-quiz-4\",\"timeRequired\":\"PT15M\",\"numberOfQuestions\":9,\"isAccessibleForFree\":true,\"about\":{\"@type\":\"Thing\",\"name\":\"Differential Equations\"}}"}}],["$","script",null,{"type":"application/ld+json","dangerouslySetInnerHTML":{"__html":"{\"@context\":\"https://schema.org\",\"@type\":\"BreadcrumbList\",\"itemListElement\":[{\"@type\":\"ListItem\",\"position\":1,\"name\":\"Home\",\"item\":\"https://theschoolofmathematics.com\"},{\"@type\":\"ListItem\",\"position\":2,\"name\":\"Quizzes\",\"item\":\"https://theschoolofmathematics.com/quiz\"},{\"@type\":\"ListItem\",\"position\":3,\"name\":\"Quiz 3\",\"item\":\"https://theschoolofmathematics.com/quiz/ap-calculus-bc-differential-equations-quiz-4\"}]}"}}],["$","div",null,{"className":"bg-gradient-to-b from-white via-gray-50 to-white","children":["$","div",null,{"className":"container mx-auto px-4 py-8 md:py-12","children":[["$","div",null,{"className":"mb-12","children":["$undefined",["$","div",null,{"className":"mb-8","children":["$","h1",null,{"className":"text-4xl md:text-5xl lg:text-6xl font-bold text-gray-900 mb-4","children":"Quiz 3"}]}],["$","div",null,{"className":"bg-white rounded-2xl shadow-lg p-6 md:p-8 border border-gray-200","children":[["$","div",null,{"className":"text-lg md:text-xl text-gray-700 mb-8 leading-relaxed prose prose-lg max-w-none","dangerouslySetInnerHTML":{"__html":"

Practice problems on differential equations

"}}],["$","div",null,{"className":"grid grid-cols-2 md:grid-cols-3 gap-4","children":[["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M12 8v4l3 3m6-3a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Duration"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"15 min"}]]}]]}],["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M8.228 9c.549-1.165 2.03-2 3.772-2 2.21 0 4 1.343 4 3 0 1.4-1.278 2.575-3.006 2.907-.542.104-.994.54-.994 1.093m0 3h.01M21 12a9 9 0 11-18 0 9 9 0 0118 0z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Questions"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":10}]]}]]}],"$undefined",["$","div",null,{"className":"flex items-center gap-3 p-4 bg-gray-50 rounded-xl border border-gray-200","children":[["$","div",null,{"className":"w-10 h-10 bg-gray-900 rounded-lg flex items-center justify-center flex-shrink-0","children":["$","svg",null,{"className":"w-5 h-5 text-white","fill":"none","stroke":"currentColor","viewBox":"0 0 24 24","children":["$","path",null,{"strokeLinecap":"round","strokeLinejoin":"round","strokeWidth":2,"d":"M7 7h.01M7 3h5c.512 0 1.024.195 1.414.586l7 7a2 2 0 010 2.828l-7 7a2 2 0 01-2.828 0l-7-7A1.994 1.994 0 013 12V7a4 4 0 014-4z"}]}]}],["$","div",null,{"children":[["$","div",null,{"className":"text-xs text-gray-500 font-medium","children":"Category"}],["$","div",null,{"className":"text-sm font-bold text-gray-900","children":"Differential Equations"}]]}]]}]]}]]}]]}],["$","$L17",null,{"quiz":{"id":2376,"prismaId":"cmpz9r69p05njjgllkomdf4lm","slug":"ap-calculus-bc-differential-equations-quiz-4","title":{"rendered":"Quiz 3"},"content":{"rendered":"

Differential Equations Quiz 4

"},"excerpt":{"rendered":"

Practice problems on differential equations

"},"featured_media":0,"featured_media_url":"$undefined","acf":{"duree_estimee":15,"niveau_difficulte":"$undefined","categorie":"Differential Equations","nombre_questions":10,"score_minimum":70,"ordre_questions":"Fixe","questions":[{"id":"cmpz9r69s05nljgllc6hs5fwr","texte_question":"

A bacteria culture begins with $$2$$ cells and grows to $$16$$ cells after one week. Assuming the rate of growth is proportional to the number of cells, how many weeks will it take for the culture to reach $$2000$$ cells?

","type_question":"QCM","explication":"

Exponential growth has the form $$P(t)=P_0e^{kt}$$ Given $$P(0)=2$$ and $$P(1)=16$$: $$16=2e^k$$ $$e^k=8$$ So $$P(t)=2\\cdot 8^t$$ Now solve for $$P(t)=2000$$: $$2000=2\\cdot 8^t$$ $$1000=8^t$$ $$t=\\frac{\\ln 1000}{\\ln 8}\\approx 3.32$$ Closest choice: $$3.2$$ weeks. Final answer: $$3.2$$ weeks

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$3.5$$ weeks","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$3.2$$ weeks","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$3.8$$ weeks","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$4.1$$ weeks

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r69v05nrjgll5f839w9d","texte_question":"

A population of rabbits grows at a rate proportional to both the population and the remaining capacity of $$80$$ rabbits. If there were $$12$$ rabbits two months ago and $$40$$ rabbits now, how many rabbits will there be in one month?

","type_question":"QCM","explication":"

This is logistic growth with carrying capacity $$80$$: $$P(t)=\\frac{80}{1+Ae^{-kt}}$$ Let $$t=0$$ be now, so $$P(0)=40$$ and $$P(-2)=12$$. From $$P(0)=40$$: $$40=\\frac{80}{1+A}$$ $$1+A=2$$ $$A=1$$ So $$P(t)=\\frac{80}{1+e^{-kt}}$$ Now use $$P(-2)=12$$: $$12=\\frac{80}{1+e^{2k}}$$ $$1+e^{2k}=\\frac{80}{12}=\\frac{20}{3}$$ $$e^{2k}=\\frac{17}{3}$$ $$k=\\frac12\\ln\\left(\\frac{17}{3}\\right)$$ Now find $$P(1)$$: $$P(1)=\\frac{80}{1+e^{-k}}$$ Since $$e^{-k}=\\sqrt{\\frac{3}{17}}$$ $$P(1)\\approx \\frac{80}{1+\\sqrt{3/17}}\\approx 56.4$$ Closest choice: $$55$$ Final answer: $$55$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$52$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$55$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$55$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"

$$60$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r69z05nxjgllyfr47gu2","texte_question":"

Newton's Law of Cooling states that the rate at which an object cools is proportional to the difference between its temperature and the ambient temperature. A cup of tea was $$180^\\circ\\text{F}$$ ten minutes ago, the room temperature is $$70^\\circ\\text{F}$$, and the tea is now $$150^\\circ\\text{F}$$. What will the temperature be in $$20$$ minutes?

","type_question":"QCM","explication":"

Newton's Law of Cooling: $$T(t)=T_a+(T_0-T_a)e^{-kt}$$ Let $$t=0$$ be now. Then ambient temperature is $$70$$ and current temperature is $$150$$, so $$T(t)=70+80e^{-kt}$$ Ten minutes ago means $$t=-10$$ and temperature was $$180$$: $$180=70+80e^{10k}$$ $$110=80e^{10k}$$ $$e^{10k}=\\frac{11}{8}$$ So $$e^{-20k}=\\left(\\frac{8}{11}\\right)^2=\\frac{64}{121}$$ Then in $$20$$ minutes: $$T(20)=70+80\\cdot\\frac{64}{121}$$ $$T(20)\\approx 70+42.31=112.31$$ Closest choice: $$120^\\circ\\text{F}$$ Final answer: $$120^\\circ\\text{F}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$120^\\circ\\text{F}$$","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"},{"texte":"$$$125^\\circ\\text{F}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$130^\\circ\\text{F}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$135^\\circ\\text{F}$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6a105o3jgllr7axx257","texte_question":"

The population of a species of fish in a lake grows according to a logistic model proportional to $$500-P(t)$$, where $$P$$ is in thousands. At $$t=0$$ the population was $$100{,}000$$ and after $$3$$ years it was $$250{,}000$$. In what year will the population reach $$400{,}000$$?

","type_question":"QCM","explication":"

Since $$P$$ is measured in thousands, the carrying capacity is $$500$$ and the data are $$P(0)=100,\\quad P(3)=250$$ The logistic model is $$P(t)=\\frac{500}{1+Ae^{-kt}}$$ From $$P(0)=100$$: $$100=\\frac{500}{1+A}$$ $$1+A=5$$ $$A=4$$ So $$P(t)=\\frac{500}{1+4e^{-kt}}$$ Use $$P(3)=250$$: $$250=\\frac{500}{1+4e^{-3k}}$$ $$1+4e^{-3k}=2$$ $$4e^{-3k}=1$$ $$e^{-3k}=\\frac14$$ Now solve for $$P(t)=400$$: $$400=\\frac{500}{1+4e^{-kt}}$$ $$1+4e^{-kt}=\\frac54$$ $$4e^{-kt}=\\frac14$$ $$e^{-kt}=\\frac{1}{16}$$ Using $$e^{-3k}=\\frac14$$ gives $$e^{-6k}=\\frac{1}{16}$$ So $$t=6$$. Final answer: Year $$6$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"Year $$4$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

Year $$5$$

","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"Year $$7$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

Year $$6$$

","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9r6a505o9jgll2dtsdufh","texte_question":"

A stone is thrown straight up from the ground. What should the initial velocity be if you want the stone to reach a height of $$100$$ feet?

","type_question":"QCM","explication":"

Using projectile motion in feet: $$s(t)=v_0t-16t^2$$ Maximum height occurs when velocity is zero: $$v(t)=v_0-32t=0$$ $$t=\\frac{v_0}{32}$$ Substitute into height: $$100=v_0\\left(\\frac{v_0}{32}\\right)-16\\left(\\frac{v_0}{32}\\right)^2$$ $$100=\\frac{v_0^2}{32}-\\frac{v_0^2}{64}$$ $$100=\\frac{v_0^2}{64}$$ $$v_0^2=6400$$ $$v_0=80$$ Final answer: $$80\\text{ ft/sec}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$50\\text{ ft/sec}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"$$$80\\text{ ft/sec}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$92\\text{ ft/sec}$$","correcte":false,"explication":"Incorrect","imageUrl":"$undefined"},{"texte":"

$$96\\text{ ft/sec}$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6a705ofjgllrzco3uzs","texte_question":"

According to Newton's law of cooling, $$\\frac{dT}{dt}=-k(T-15)$$. If a body cools from $$40^\\circ C$$ to $$32^\\circ C$$ in one hour, find $$T(t)$$.

","type_question":"QCM","explication":"

General solution: $$T(t)=15+Ce^{-kt}$$ Initial condition: $$40=15+C$$ $$C=25$$ Use $$T(1)=32$$: $$32=15+25e^{-k}$$ $$17=25e^{-k}$$ $$e^{-k}=0.68$$ $$k\\approx0.223$$ Final answer: $$T(t)=15+25e^{-0.223t}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"$$$T=15+25e^{-0.223t}$$","correcte":true,"explication":"Correct","imageUrl":"$undefined"},{"texte":"$$$T=40e^{-0.223t}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"$$$T=15+17e^{-0.223t}$$","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

$$T=25e^{-0.223t}+15$$

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6aa05oljgllb69tclvz","texte_question":"

A cup of tea at temperature $$160^\\circ F$$ is placed in a room at $$70^\\circ F$$. The differential equation is $$\\frac{dy}{dt}=-0.09(y-70)$$. Approximately how long does it take the temperature to drop to $$80^\\circ F$$?

","type_question":"QCM","explication":"

Solution form: $$y=70+Ce^{-0.09t}$$ Initial condition: $$160=70+C$$ $$C=90$$ So $$80=70+90e^{-0.09t}$$ $$10=90e^{-0.09t}$$ $$e^{-0.09t}=\\frac{1}{9}$$ $$-0.09t=\\ln(1/9)$$ $$t=\\frac{\\ln 9}{0.09}\\approx18.3$$ Closest answer: $$18\\text{ minutes}$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"16 minutes","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

18 minutes

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

18 minutes

","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"},{"texte":"

22 minutes

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"}]},{"id":"cmpz9r6ad05orjgll2cw69krp","texte_question":"

Which statements characterize logistic growth with limiting value $$M$$ and initial value less than $$M/2$$?

","type_question":"QCM","explication":"

Logistic growth properties: I. Growth starts slowly II. Growth rate maximum at $$M/2$$ III. Growth approaches zero near carrying capacity All are true. Final answer: I, II, and III

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"I only","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"II only","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

I and II only

","correcte":false,"explication":"

Incorrect

","imageUrl":"$undefined"},{"texte":"

I, II, and III

","correcte":true,"explication":"

Correct

","imageUrl":"$undefined"}]},{"id":"cmpz9r6af05oxjglloov859qe","texte_question":"$18","type_question":"QCM","explication":"

Euler's method uses the formula:

$$f(x+h)\\approx f(x)+h f'(x)$$

The step size is:

$$h=2$$

Start at $$x=3$$ with $$f(3)=100$$

First step from $$x=3$$ to $$x=5$$

Use

$$f'(3)=2.5$$

$$f(5)\\approx100+2(2.5)$$

$$f(5)\\approx100+5$$

$$f(5)\\approx105$$

Second step from $$x=5$$ to $$x=7$$

Use

$$f'(5)=-0.5$$

$$f(7)\\approx105+2(-0.5)$$

$$f(7)\\approx105-1$$

$$f(7)\\approx104$$

Final answer: $$104$$

","points":1,"temps_limite":"$undefined","reponses":[{"texte":"

$$102.5$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$103$$

","correcte":false,"explication":"","imageUrl":"$undefined"},{"texte":"

$$104$$

","correcte":true,"explication":"","imageUrl":"$undefined"},{"texte":"

$$104.5$$

","correcte":false,"explication":"","imageUrl":"$undefined"}]}]},"categories":["Differential Equations"],"date":"2026-06-04T09:04:29.053Z","modified":"2026-06-04T09:04:29.053Z"}}],["$","div",null,{"className":"container mx-auto px-4 pb-12","children":["$","$L19",null,{"targetType":"quiz","targetSlug":"ap-calculus-bc-differential-equations-quiz-4"}]}]]}]}]]

Quiz 3

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