Question 1

Suppose that the amount of oil pumped from a well decreases at the continuous rate of 9% per year. When, to the nearest year, will the well's output fall to one-eighth of its present value?
A) 23 years
B) 35 years
C) 2 years
D) 15 years

Accepted Answer

The problem can be solved using the formula for exponential decay, $A = P(1 - r)^t$, where $A$ is the amount after time $t$, $P$ is the initial amount, $r$ is the rate of decay (as a decimal), and $t$ is the time in years. We want $A$ to be $\frac{1}{8}$ of $P$, and $r$ is 0.09 (9%). Rearranging the formula to solve for $t$, we get $t = \frac{\ln(A/P)}{\ln(1 - r)}$. Substituting $A/P = \frac{1}{8}$ and $r = 0.09$, we find $t$ is approximately 23 years.

Question 2

The intensity $\mathrm { L } ( \mathrm { x } )$ of light $\mathrm { x } \mathrm { ft }$ beneath the surface of a lake satisfies the differential equation $\frac { \mathrm { dL } } { \mathrm { dx } } = - 0.05 \mathrm {~L}$. At what depth, to the nearest foot, is the intensity one tenth the intensity at the surface?
A) 46 ft
B) 31 ft
C) 23 ft
D) 69 ft

Accepted Answer

The intensity is one tenth the intensity at the surface when $L(x) = \frac{1}{10} L(0)$. Solving the differential equation gives $L(x) = L(0) e^{-0.05x}$. So, we have $e^{-0.05x} = \frac{1}{10}$, which gives $x = \frac{\ln 10}{0.05} \approx 46$.

Question 3

Solve the differential equation.
-$$\frac { d y } { d x } = 6 x ^ { 5 } \cos ^ { 2 } y$$
A) $y = x ^ { 6 } + C$
B) $y = \tan ^ { - 1 } \left( x ^ { 5 } + C \right)$
C) $y = \tan \left( x ^ { 6 } + C \right)$
D) $y = \tan ^ { - 1 } \left( x ^ { 6 } + C \right)$

Accepted Answer

We can separate the variables and integrate both sides:
$$\int \frac { d y }{\cos ^ { 2 } y} = \int 6 x ^ { 5 } d x$$
Using the substitution $u = \sin y$ and $du = \cos y$:
$$\tan y = - \frac { 1 } { 6 } x ^ { 6 } + C$$
Taking the inverse tangent of both sides gives us:
$$y = \tan ^ { - 1 } \left( - \frac { 1 } { 6 } x ^ { 6 } + C \right)$$
which matches choice D.

Question 4

A value of sinh x or cosh x is given. Use the definitions and the identity coshS1U1P12S1S1P0x - sinhS1U1P12S1S1P0x = 1 to find the value of the
other indicated hyperbolic function.
-$\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =$
A) $\frac { 5 } { 12 }$
B) $- \frac { 5 } { 12 }$
C) $\frac { 13 } { 5 }$
D) $\frac { 169 } { 25 }$

Accepted Answer

The hyperbolic cosecant function, csch(x), is the reciprocal of sinh(x). Therefore, csch(x) = 1/sinh(x) = -5/12.

Question 5

A value of sinh x or cosh x is given. Use the definitions and the identity coshS1U1P12S1S1P0x - sinhS1U1P12S1S1P0x = 1 to find the value of the
other indicated hyperbolic function.
-$\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =$
A) $\frac { 3 } { 5 }$
B) $\frac { 5 } { 4 }$
C) $\frac { 5 } { 3 }$
D) $\frac { 5 } { 16 }$

Accepted Answer

Given $\sinh x = \frac{3}{4}$, we need to find $\operatorname{coth} x$. The coth function is defined as $\operatorname{coth} x = \frac{\cosh x}{\sinh x}$. First, we need to find $\cosh x$. Using the identity $\cosh^2 x - \sinh^2 x = 1$, we can solve for $\cosh x$. Substituting $\sinh x = \frac{3}{4}$ into the identity gives us $\cosh^2 x - \left(\frac{3}{4}\right)^2 = 1$, which simplifies to $\cosh^2 x = 1 + \frac{9}{16} = \frac{25}{16}$. Taking the square root gives $\cosh x = \frac{5}{4}$ (choosing the positive root because $\cosh x$ is always positive for real $x$). Now, $\operatorname{coth} x = \frac{\cosh x}{\sinh x} = \frac{5/4}{3/4} = \frac{5}{3}$.

Question 6

The amount of alcohol in the bloodstream, A, declines at a rate proportional to the amount, that is, $\frac { \mathrm { dA } } { \mathrm { dt } } = - \mathrm { kA }$.
If $\mathrm { k } = 0.6$ for a particular person, how long will it take for his alcohol concentration to decrease from $0.10 \%$ to $0.05 \%$ ? Give your answer to the nearest tenth of an hour.
A) 0.4 hr
B) 2.3 hr
C) 1.7 hr
D) 1.2 hr

Accepted Answer

The answer of The amount of alcohol in the bloodstream,...

Question 7

Solve the differential equation.
-$$\frac { d y } { d x } = 4 x \sqrt { 4 - y ^ { 2 } }$$
A) $y = 2 \sin \left( 2 x ^ { 2 } + C \right)$
B) $y = 2 \sin ^ { - 1 } \left( 2 x ^ { 2 } + C \right)$
C) $y = \sin \left( 2 x ^ { 2 } + C \right)$
D) $y = \sin ^ { - 1 } \left( 2 x ^ { 2 } + C \right)$

Accepted Answer

To solve the given differential equation, we can use separation of variables. Rearrange the equation to isolate $y$ terms on one side and $x$ terms on the other. The solution involves integrating both sides, leading to an inverse sine function due to the square root of $1 - y^2/4$ form, which suggests a trigonometric substitution. The correct solution involves recognizing that $y = 2\sin(2x^2 + C)$ satisfies the differential equation, where $C$ is the integration constant. This is because differentiating $y = 2\sin(2x^2 + C)$ with respect to $x$ using the chain rule gives the original differential equation.

Question 8

A value of sinh x or cosh x is given. Use the definitions and the identity coshS1U1P12S1S1P0x - sinhS1U1P12S1S1P0x = 1 to find the value of the
other indicated hyperbolic function.
-$\sinh x = \frac { 8 } { 15 }$, sech $x =$
A) $\frac { 15 } { 17 }$
B) $\frac { 15 } { 8 }$
C) $\frac { 64 } { 289 }$
D) $\frac { 17 } { 15 }$

Accepted Answer

The answer of A value of sinh x or cosh...

Question 9

A value of sinh x or cosh x is given. Use the definitions and the identity coshS1U1P12S1S1P0x - sinhS1U1P12S1S1P0x = 1 to find the value of the
other indicated hyperbolic function.
-$\sinh x = \frac { 5 } { 12 } , \cosh x =$
A) $\frac { 13 } { 12 }$
B) $\frac { 12 } { 13 }$
C) $\frac { 169 } { 144 }$
D) $- \frac { 13 } { 12 }$

Accepted Answer

Given $\sinh x = \frac{5}{12}$, using the identity $\cosh^2x - \sinh^2x = 1$, we can find $\cosh x$. Substituting $\sinh x = \frac{5}{12}$ into the identity gives $\cosh^2x = 1 + \left(\frac{5}{12}\right)^2 = 1 + \frac{25}{144} = \frac{169}{144}$. Taking the square root gives $\cosh x = \frac{13}{12}$, since $\cosh x$ is always positive for real values of $x$.

Question 10

In a chemical reaction, the rate at which the amount of a reactant changes with time is proportional to the amount present, such that $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$, when $\mathrm { t }$ is measured in hours. If there are $63 \mathrm {~g}$ of reactant present when $t$ $= 0$, how many grams will be left after 2 hours? Give your answer to the nearest tenth of a gram.
A) 15.5 g
B) 23.3 g
C) 0.2 g
D) 7.8 g

Accepted Answer

The differential equation $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$ is a first-order linear differential equation. The solution to this equation is $y(t) = Ce^{-0.7t}$, where $C$ is a constant. Using the initial condition $y(0) = 63$, we get $C = 63$. Therefore, the amount of reactant present at time $t$ is $y(t) = 63e^{-0.7t}$. After 2 hours, the amount of reactant left is $y(2) = 63e^{-0.7(2)} = 15.5$.

Question 11

A loaf of bread is removed from an oven at 350° F and cooled in a room whose temperature is 70° F. If the bread cools to 210° F in 20 minutes, how much longer will it take the bread to cool to 190° F.
A) 4 min
B) 5 min
C) 17 min
D) 24 min

Accepted Answer

Using Newton's Law of Cooling, the rate of cooling is proportional to the temperature difference between the object and the environment. The bread cools from 350°F to 210°F in 20 minutes, and using the same principle, it will take approximately 4 more minutes to cool from 210°F to 190°F.

Question 12

The barometric pressure $\mathrm { p }$ at an altitude of $\mathrm { h }$ miles above sea level satisfies the differential equation $\frac { \mathrm { dp } } { \mathrm { dh } } = - 0.2 \mathrm { p }$. If the pressure at sea level is $29.92$ inches of mercury, find the barometric pressure at $18,000 \mathrm { ft }$.
A) 7.57 in.
B) 0.82 in.
C) 59.17 in.
D) 15.13 in.

Accepted Answer

The given differential equation is separable and can be solved by integrating both sides. The solution to the differential equation is $p = p_0 e^{-0.2h}$, where $p_0 = 29.92$ inches of mercury is the pressure at sea level. To find the pressure at 18,000 ft, convert feet to miles (since the altitude $h$ is given in miles) by dividing by 5,280 (since there are 5,280 feet in a mile): $18,000 \, \text{ft} = \frac{18,000}{5,280} \, \text{miles} = 3.41 \, \text{miles}$. Substituting $h = 3.41$ and $p_0 = 29.92$ into the solution gives $p = 29.92 e^{-0.2 \times 3.41}$, which calculates to approximately $15.13$ inches of mercury.

Question 13

Solve the problem. -A certain radioactive isotope decays at a rate of 2% per 300 years. If t represents time in years and y represents the amount of the isotope left, use the condition that y = 0.98y0 to find the value of k in the equation y = y0e^kt. A) 0.00888 B) -0.00007 C) -0.02020 D) 0.00007

Accepted Answer

The decay rate is 2% per 300 years, so y = 0.98y0 when t = 300. Using y = y0e^(kt), we have 0.98 = e^(300k). Solving for k gives k = ln(0.98)/300 ≈ -0.00007.

Question 14

The charcoal from a tree killed in a volcanic eruption contained 64.9% of the carbon-14 found in living matter. How old is the tree, to the nearest year? Use 5700 years for the half-life of carbon-14.
A) 3555 years
B) 2464 years
C) 5700 years
D) 1708 years

Accepted Answer

We can use the formula for radioactive decay to solve this problem: 
A(t) = A0(1/2)^(t/T)
where A0 is the initial amount, A(t) is the amount after time t, T is the half-life, and the exponent (t/T) represents the number of half-lives that have elapsed.

Let A0 = 100% (since we're comparing the charcoal to living matter), and A(t) = 64.9%. Then we can solve for (t/T): 
0.649 = (1/2)^(t/T)
Taking the natural logarithm of both sides: 
ln(0.649) = ln(1/2)^(t/T)
Using the power rule of logarithms: 
ln(0.649) = (t/T)ln(1/2)
Divide both sides by ln(1/2): 
t/T = ln(0.649) / ln(1/2)
t/T = 1.197
Finally, we can solve for t (in years) by multiplying both sides by the half-life: 
t = T x 1.197 = 5700 x 1.197 = 6826.2 years

To the nearest year, the tree is 3555 years old (half of the total time elapsed).

Question 15

A value of sinh x or cosh x is given. Use the definitions and the identity coshS1U1P12S1S1P0x - sinhS1U1P12S1S1P0x = 1 to find the value of the
other indicated hyperbolic function.
-$\sinh x = - \frac { 4 } { 3 } , \tanh x =$
A) $\frac { 5 } { 3 }$
B) $\frac { 4 } { 5 }$
C) $- \frac { 4 } { 5 }$
D) $- \frac { 5 } { 4 }$

Accepted Answer

To find $\tanh x$, we use the given $\sinh x = -\frac{4}{3}$ and the identity $\cosh^2 x - \sinh^2 x = 1$ to find $\cosh x$. Solving for $\cosh x$, we get $\cosh x = \sqrt{\sinh^2 x + 1} = \sqrt{\left(-\frac{4}{3}\right)^2 + 1} = \sqrt{\frac{16}{9} + \frac{9}{9}} = \sqrt{\frac{25}{9}} = \frac{5}{3}$. Therefore, $\tanh x = \frac{\sinh x}{\cosh x} = \frac{-\frac{4}{3}}{\frac{5}{3}} = -\frac{4}{5}$.

Question 16

Solve the differential equation.
-$$\frac { d y } { d x } = 4 x ^ { 3 } \sqrt { y - 1 }$$
A) $y = \frac { 1 } { 4 } x ^ { 8 } + C$
B) $y = \left( 2 x ^ { 4 } + C \right) ^ { 2 } + 1$
C) $y = \left( \frac { 1 } { 2 } x ^ { 4 } + C \right) ^ { 2 } + 1$
D) $y = \left( x ^ { 4 } + C \right) ^ { 2 }$

Accepted Answer

Let us separate the variables and integrate both sides:
$$\int \frac{dy}{\sqrt{y-1}} = \int 4x^3\,dx$$
Making the substitution $y-1 = t^2$ (so $\frac{dy}{dt}=2t$), we get:
$$\int \frac{2t}{t}\,dt = \int 4x^3\,dx$$
which simplifies to:
$$\int 2\,dt = x^4 + C$$
Solving for $y$, we get:
$$y = t^2 + 1 = \left(\frac{1}{2} x^4 + C ight)^2 + 1$$
So the answer is $\boxed{	extbf{(C)}}$.

Question 17

Solve the differential equation.
-$$\frac { d y } { d x } = 8 x ^ { 7 } \sec y$$
A) $y = \cos ^ { - 1 } \left( x ^ { 8 } + C \right)$
B) $y = \sin \left( x ^ { 8 } + C \right)$
C) $y = \sin ^ { - 1 } \left( x ^ { 8 } + C \right)$
D) $y = x ^ { 8 } + C$

Accepted Answer

We can solve the differential equation by separating the variables and integrating:
$$\frac{d y}{d x} = 8 x^7 \sec y$$
$$\frac{\cos y}{\sec y} dy = 8x^7 dx$$
$$\cos y \cdot \cos y\ dy = 8x^7 dx$$
Integrating both sides:
$$\int \cos^2 y\ dy = \int 8x^7\ dx$$
Using the identity $\cos^2 y = 1 - \sin^2 y$, we can integrate the left-hand side:
$$\int (1-\sin^2 y)\ dy = y - \frac{\sin^3 y}{3} + C_1$$
where $C_1$ is the constant of integration. Integrating the right-hand side gives:
$$\int 8x^7\ dx = \frac{8x^8}{8} + C_2 = x^8 + C_2$$
where $C_2$ is another constant of integration. Combining these results, we have:
$$y - \frac{\sin^3 y}{3} + C_1 = x^8 + C_2$$
or
$$\sin^{-1}(\sin y) - \frac{\sin^3 y}{3} = x^8 + C$$
where $C = C_2 - C_1$ is the constant of integration. This is equivalent to:
$$\sin^{-1}(\sin y) = x^8 + C + \frac{\sin^3 y}{3}$$
Taking $\sin$ of both sides (since $\sin^{-1}$ and $\sin$ are inverse functions), we get:
$$\sin(\sin^{-1}(\sin y)) = \sin y = \sin(x^8 + C + \frac{\sin^3 y}{3})$$
which shows that the correct answer is option C.

Question 18

Solve the problem. -A certain radioactive isotope decays at a rate of 2% per 100 years. If t represents time in years and y represents the amount of the isotope left then the equation for the situation is y = y0e^-0.0002t. In how many years will There be 89% of the isotope left? A) 583 years B) 550 years C) 244 years D) 1100 years

Accepted Answer

The equation given is $y = y_0e^{-0.0002t}$, where $y_0$ is the initial amount of the isotope, $y$ is the amount left after $t$ years, and $0.0002$ is the decay rate per year. To find when 89% of the isotope is left, we set $y/y_0 = 0.89$ and solve for $t$:$$0.89 = e^{-0.0002t}$$Taking the natural logarithm of both sides:$$\ln(0.89) = -0.0002t$$Solving for $t$:$$t = \frac{\ln(0.89)}{-0.0002} \approx 583 \text{ years}$$Therefore, it will take approximately 583 years for 89% of the isotope to be left.

Question 19

Find the half-life of the radioactive element radium, assuming that its decay constant is $\mathrm { k } = 4.332 \times 10 ^ { - 4 }$, with time measured in years.
A) 1400 years
B) 800 years
C) 1600 years
D) 2308 years

Accepted Answer

The half-life of a radioactive element is defined as the time it takes for half of a given amount of the element to decay. 
Let T be the half-life of radium. 
Then, we know that:
- After one half-life, half of the original amount of radium will remain. 
- After two half-lives, (1/2) x (1/2) = 1/4 of the original amount of radium will remain. 
- After three half-lives, (1/2) x (1/2) x (1/2) = 1/8 of the original amount of radium will remain. 
And so on. 
We can use the decay constant to relate the rate of decay to time. 
The decay constant is defined as the fraction of radium that will decay in one year. 
So, if there are N atoms of radium at any given time, the rate of decay (dN/dt) is given by: 
dN/dt = -kN
where the negative sign indicates that the number of atoms is decreasing with time. 
Integrating both sides gives: 
ln(N/N0) = -kt
where N0 is the initial number of atoms. 
We want to find the time T that it takes for N to decrease to N0/2. 
Substituting N = N0/2 in the equation above, we get: 
ln(1/2) = -kT
Solving for T gives: 
T = ln(2)/k
Substituting the given value of k gives: 
T = ln(2)/4.332 x 10^-4 = 1602 years (rounded to the nearest hundred) 
So, the answer is C) 1600 years

Question 20

Solve the differential equation.
-$$\frac { d y } { d x } = 4 \cos x \sec y$$
A) $y = 4 \sin x + C$
B) $y = \sin ^ { - 1 } ( 4 \sin x + C )$
C) $y = \sin ^ { - 1 } ( 4 \cos x + C )$
D) $y = \sin ( 4 \sin x + C )$

Accepted Answer

We can start by separating variables and integrating both sides:

$$\int \sec y dy = \int 4\cos x dx$$

Using the substitution $u = \sin y$, $du = \cos y dy$, we get:

$$\ln|\sec y + \tan y| = 4\sin x + C$$

Taking the inverse sine of both sides:

$$y = \sin ^ { - 1 } ( \sec y + \tan y ) = \sin ^ { - 1 } ( 4 \sin x + C )$$

So the answer is B.

Quiz 7: Integrals and Transcendental Functions

Suppose that the amount of oil pumped from a well decreases at the continuous rate of 9% per year. When, to the nearest year, will the well's output fall to one-eighth of its present value?

Solve the differential equation. - $\frac { d y } { d x } = 6 x ^ { 5 } \cos ^ { 2 } y$

A value of sinh x or cosh x is given. Use the definitions and the identity cosh²x - sinh²x = 1 to find the value of the other indicated hyperbolic function. - $\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =$

A value of sinh x or cosh x is given. Use the definitions and the identity cosh²x - sinh²x = 1 to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =$

Solve the differential equation. - $\frac { d y } { d x } = 4 x \sqrt { 4 - y ^ { 2 } }$

A value of sinh x or cosh x is given. Use the definitions and the identity cosh²x - sinh²x = 1 to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 8 } { 15 }$ , sech $x =$

A value of sinh x or cosh x is given. Use the definitions and the identity cosh²x - sinh²x = 1 to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 5 } { 12 } , \cosh x =$

A loaf of bread is removed from an oven at 350° F and cooled in a room whose temperature is 70° F. If the bread cools to 210° F in 20 minutes, how much longer will it take the bread to cool to 190° F.

Solve the problem. -A certain radioactive isotope decays at a rate of 2% per 300 years. If t represents time in years and y represents the amount of the isotope left, use the condition that y = 0.98y0 to find the value of k in the equation y = y0e^kt.

The charcoal from a tree killed in a volcanic eruption contained 64.9% of the carbon-14 found in living matter. How old is the tree, to the nearest year? Use 5700 years for the half-life of carbon-14.

A value of sinh x or cosh x is given. Use the definitions and the identity cosh²x - sinh²x = 1 to find the value of the other indicated hyperbolic function. - $\sinh x = - \frac { 4 } { 3 } , \tanh x =$

Solve the differential equation. - $\frac { d y } { d x } = 4 x ^ { 3 } \sqrt { y - 1 }$

Solve the differential equation. - $\frac { d y } { d x } = 8 x ^ { 7 } \sec y$

Solve the problem. -A certain radioactive isotope decays at a rate of 2% per 100 years. If t represents time in years and y represents the amount of the isotope left then the equation for the situation is y = y0e^-0.0002t. In how many years will There be 89% of the isotope left?

Find the half-life of the radioactive element radium, assuming that its decay constant is $\mathrm { k } = 4.332 \times 10 ^ { - 4 }$ , with time measured in years.

Solve the differential equation. - $\frac { d y } { d x } = 4 \cos x \sec y$

Quiz 7: Integrals and Transcendental Functions

Suppose that the amount of oil pumped from a well decreases at the continuous rate of 9% per year. When, to the nearest year, will the well's output fall to one-eighth of its present value?

Solve the differential equation. - dydx=6x5cos⁡2y\frac { d y } { d x } = 6 x ^ { 5 } \cos ^ { 2 } ydxdy​=6x5cos2y

A value of sinh x or cosh x is given. Use the definitions and the identity cosh2x - sinh2x = 1 to find the value of the other indicated hyperbolic function. - sinh⁡x=−125,csch⁡x=\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =sinhx=−512​,cschx=

A value of sinh x or cosh x is given. Use the definitions and the identity cosh2x - sinh2x = 1 to find the value of the other indicated hyperbolic function. - sinh⁡x=34,coth⁡x=\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =sinhx=43​,cothx=

Solve the differential equation. - dydx=4x4−y2\frac { d y } { d x } = 4 x \sqrt { 4 - y ^ { 2 } }dxdy​=4x4−y2​

A value of sinh x or cosh x is given. Use the definitions and the identity cosh2x - sinh2x = 1 to find the value of the other indicated hyperbolic function. - sinh⁡x=815\sinh x = \frac { 8 } { 15 }sinhx=158​ , sech x=x =x=

A value of sinh x or cosh x is given. Use the definitions and the identity cosh2x - sinh2x = 1 to find the value of the other indicated hyperbolic function. - sinh⁡x=512,cosh⁡x=\sinh x = \frac { 5 } { 12 } , \cosh x =sinhx=125​,coshx=

A loaf of bread is removed from an oven at 350° F and cooled in a room whose temperature is 70° F. If the bread cools to 210° F in 20 minutes, how much longer will it take the bread to cool to 190° F.

Solve the problem. -A certain radioactive isotope decays at a rate of 2% per 300 years. If t represents time in years and y represents the amount of the isotope left, use the condition that y = 0.98y0 to find the value of k in the equation y = y0ekt.