Question 1

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function.
-$\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =$
A) $\frac { 169 } { 25 }$
B) $\frac { 5 } { 12 }$
C) $\frac { 13 } { 5 }$
D) $- \frac { 5 } { 12 }$

Accepted Answer

The hyperbolic cosecant, $\operatorname{csch} x$, is the reciprocal of $\sinh x$. Therefore, $\operatorname{csch} x = \frac{1}{\sinh x} = -\frac{5}{12}$.

Question 2

Solve the problem.
-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) 800 years
B) 1400 years
C) 2308 years
D) 1600 years

Accepted Answer

The answer of Solve the problem.
-Find the half-life of the...

Question 3

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

Accepted Answer

We have $\frac { d y } { d x } = 8 x ^ { 7 } \sec y$. Integrating both sides with respect to $x$ we get $\int\frac {1}{\sec y}dy =  \int 8 x ^ { 7 } dx$. The left-hand side can be simplified using the identity $\sec^2 y=1+	an^2y$. Thus, we have $\ln| \sec y + 	an y | = x^8+C$ where $C$ is the constant of integration. Taking the inverse sine of both sides, we get $\sin^{-1}(\sec y + 	an y) = x^8+C$, which can be rearranged to obtain $y = \sin^{-1}(x^8+C)$. Therefore, the best choice is $\boxed{	ext{(A) } y = \sin ^ { - 1 } \left( x ^ { 8 } + C ight)}$.

Question 4

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function.
-$\sinh x = \frac { 8 } { 15 } , \operatorname { sech } x =$
A) $\frac { 17 } { 15 }$
B) $\frac { 15 } { 8 }$
C) $\frac { 64 } { 289 }$
D) $\frac { 15 } { 17 }$

Accepted Answer

Given $\sinh x = \frac{8}{15}$, we use the identity $\cosh^2 x - \sinh^2 x = 1$ to find $\cosh x$. First, solve for $\cosh x$: $\cosh^2 x = \sinh^2 x + 1 = \left(\frac{8}{15}\right)^2 + 1 = \frac{64}{225} + 1 = \frac{289}{225}$. Thus, $\cosh x = \sqrt{\frac{289}{225}} = \frac{17}{15}$. The $\operatorname{sech} x$ is the reciprocal of $\cosh x$, so $\operatorname{sech} x = \frac{1}{\cosh x} = \frac{15}{17}$.

Question 5

Solve the problem.
-Suppose that the amount of oil pumped from a well decreases at the continuous rate of 12% per year. When, to the nearest year, will the well's output fall to one-eighth of its present value?
A) 17 years
B) 12 years
C) 26 years
D) 2 years

Accepted Answer

The problem involves exponential decay. The formula for exponential decay is $ A = P e^{-rt} $, where $ A $ is the final amount, $ P $ is the initial amount, $ r $ is the rate of decay, and $ t $ is time. We need to find $ t $ when $ A = \frac{1}{8}P $ and $ r = 0.12 $. Solving $ \frac{1}{8}P = P e^{-0.12t} $ gives $ t \approx 17 $ years.

Question 6

Solve the problem.
-A certain radioactive isotope decays at a rate of $1 \%$ per 400 years. If $t$ represents time in years and y represents the amount of the isotope left, use the condition that $\mathrm { y } = 0.99 \mathrm { y } 0$ to find the value of $\mathrm { k }$ in the equation $\mathrm { y } = \mathrm { y } _ { 0 } \mathrm { e } ^ { \mathrm { kt } }$.
A) 0.00003
B) -0.00003
C) -0.01005
D) 0.00673

Accepted Answer

The decay of a radioactive isotope can be modeled by the equation $y = y_0 e^{kt}$, where $y$ is the amount of the isotope remaining, $y_0$ is the initial amount, $k$ is the decay constant, and $t$ is time. Given that the isotope decays by $1\%$ in 400 years, we have $y = 0.99y_0$ when $t = 400$. Substituting these values into the decay equation gives $0.99y_0 = y_0 e^{400k}$. Dividing both sides by $y_0$ and taking the natural logarithm of both sides gives $\ln(0.99) = 400k$. Solving for $k$ yields $k = \frac{\ln(0.99)}{400}$, which is approximately $-0.000025$, but the closest option provided is $-0.00003$.

Question 7

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 $y = y 0 e ^ { - 0.0002 t }$. In how many years will there be $89 \%$ of the isotope left?
A) 550 years
B) 583 years
C) 1100 years
D) 244 years

Accepted Answer

We want to solve for $t$ when $y = 0.89y_0$.
$$0.89y_0 = y_0 e^{-0.0002t}$$
Dividing both sides by $y_0$ gives
$$0.89 = e^{-0.0002t}$$
Taking the natural logarithm of both sides gives
$$\ln(0.89) = -0.0002t\ln(e) = -0.0002t$$
Solving for $t$ gives
$$t = \frac{\ln(0.89)}{-0.0002} \approx 582.67$$
Therefore, the answer is B) 583 years (rounded to the nearest year).

Question 8

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

Accepted Answer

To solve the differential equation $\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$, we separate variables and integrate both sides. Rearranging gives $\frac{dy}{\cos^2 y} = 3x^2 dx$. Integrating both sides, we get $\tan^{-1}(y) = x^3 + C$, which implies $y = \tan^{-1}(x^3 + C)$.

Question 9

Solve the differential equation.
-$$\frac { d y } { d x } = 3 \cos x \sec y$$
A) $y = 3 \sin x + C$
B) $y = \sin ^ { - 1 } ( 3 \cos x + C )$
C) $y = \sin ( 3 \sin x + C )$
D) $y = \sin ^ { - 1 } ( 3 \sin x + C )$

Accepted Answer

We can separate the variables and write:
$$\sec y dy = 3 \cos x dx$$
Taking the antiderivative of both sides, we get:
$$\ln|\sec y + \tan y| = 3\sin x + C_1$$
where $C_1$ is the constant of integration.

Applying $\sin^{-1}$ to both sides, we have:
$$y = \sin^{-1}(\sec y + \tan y) = \sin^{-1}(3\cos x + C)$$
where $C = \sin^{-1}(C_1)$. Therefore, the answer is $\boxed{\textbf{(D)}}$.

Question 10

Solve the problem.
-The intensity $\mathrm { L } ( \mathrm { x } )$ of light $x \mathrm { ft }$ beneath the surface of a lake satisfies the differential equation $\frac { \mathrm { dL } } { \mathrm { dx } } = - 0.07 \mathrm {~L} .$ At what depth, to the nearest foot, is the intensity one tenth the intensity at the surface?
A) 22 ft
B) 49 ft
C) 16 ft
D) 33 ft

Accepted Answer

We can solve this separable differential equation by dividing both sides by L and integrating:
$$\ln L = -0.07x + C$$
where C is the constant of integration. Exponentiating both sides, we get:
$$L = e^{-0.07x+C} = Ce^{-0.07x}$$
where we have combined the constant of integration C into a single constant.

To find the depth at which the intensity is one tenth the surface intensity, we can use the fact that:
$$\frac{L(d)}{L(0)} = \frac{1}{10}$$
where d is the depth we want to find. Substituting our expression for L, we get:
$$\frac{Ce^{-0.07d}}{C} = \frac{1}{10}$$
Simplifying, we get:
$$e^{-0.07d} = \frac{1}{10}$$
Taking the natural logarithm of both sides, we get:
$$-0.07d = \ln\frac{1}{10} = -\ln 10$$
Solving for d, we get:
$$d = \frac{\ln 10}{0.07} \approx 33$$
to the nearest foot. Therefore, the answer is D, 33 ft.

Question 11

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function.
-$\sinh x = - \frac { 4 } { 3 } , \tanh x =$
A) $- \frac { 5 } { 4 }$
B) $\frac { 5 } { 3 }$
C) $- \frac { 4 } { 5 }$
D) $\frac { 4 } { 5 }$

Accepted Answer

Given $\sinh x = -\frac{4}{3}$, we can find $\cosh x$ using the identity $\cosh^2 x - \sinh^2 x = 1$. Substituting $\sinh x = -\frac{4}{3}$ into the identity gives us $\cosh^2 x - \left(-\frac{4}{3}\right)^2 = 1$, which simplifies to $\cosh^2 x = 1 + \frac{16}{9}$, leading to $\cosh^2 x = \frac{25}{9}$. Taking the square root gives $\cosh x = \frac{5}{3}$ (choosing the positive root because $\cosh x$ is always positive). Then, $\tanh x = \frac{\sinh x}{\cosh x} = \frac{-\frac{4}{3}}{\frac{5}{3}} = -\frac{4}{5}$.

Question 12

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function.
-$\sinh x = \frac { 5 } { 12 } , \cosh x =$
A) $\frac { 12 } { 13 }$
B) $\frac { 169 } { 144 }$
C) $\frac { 13 } { 12 }$
D) $- \frac { 13 } { 12 }$

Accepted Answer

The answer of A value of $\sinh x$ or \(\cosh...

Question 13

Solve the differential equation.
-$$\frac { d y } { d x } = 6 x \sqrt { 9 - y ^ { 2 } }$$
A) $y = 3 \sin \left( 3 x ^ { 2 } + C \right)$
B) $y = 3 \sin ^ { - 1 } \left( 3 x ^ { 2 } + C \right)$
C) $y = \sin ^ { - 1 } \left( 3 x ^ { 2 } + C \right)$
D) $y = \sin \left( 3 x ^ { 2 } + C \right)$

Accepted Answer

This differential equation can be solved by separation of variables. Rearranging, we get $\frac{dy}{\sqrt{9-y^2}} = 6x dx$. Integrating both sides, we use the substitution $y = 3\sin(u)$ for the left side, leading to $u = 3x^2 + C$. Thus, $y = 3\sin(3x^2 + C)$ is the solution.

Question 14

Solve the problem.
-A loaf of bread is removed from an oven at $350 ^ { \circ } \mathrm { F }$ and cooled in a room whose temperature is $70 ^ { \circ } \mathrm { F }$. If the bread cools to $210 ^ { \circ } \mathrm { F }$ in 20 minutes, how much longer will it take the bread to cool to $185 ^ { \circ } \mathrm { F }$.
A) 18 min
B) 7 min
C) 6 min
D) 26 min

Accepted Answer

Using Newton's Law of Cooling, the time to cool from 210°F to 185°F is approximately 6 minutes.

Question 15

Solve the problem.
-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 $78 \mathrm {~g}$ of reactant present when $t$ $= 0$, how many grams will be left after 4 hours? Give your answer to the nearest tenth of a gram.
A) 4.7 g
B) 2.4 g
C) 0.1 g
D) 7.1 g

Accepted Answer

The given differential equation $\frac { \mathrm { dy } } { \mathrm { dt } } = - 0.7 \mathrm { y }$ is a first-order linear differential equation, which can be solved to find the amount of reactant $y$ at any time $t$. The solution to this equation is $y = y_0 e^{-0.7t}$, where $y_0$ is the initial amount of the reactant. Given $y_0 = 78$ g and $t = 4$ hours, the amount of reactant left after 4 hours is $y = 78e^{-0.7 \times 4}$. Calculating this gives approximately $4.7$ g.

Question 16

Solve the differential equation.
-$$\frac { d y } { d x } = 8 x ^ { 7 } \sqrt { y - 1 }$$
A) $y = \left( 2 x ^ { 8 } + C \right) ^ { 2 } + 1$
B) $y = \frac { 1 } { 4 } x ^ { 16 } + C$
C) $y = \left( \frac { 1 } { 2 } x ^ { 8 } + C \right) ^ { 2 } + 1$
D) $y = \left( x ^ { 8 } + C \right) ^ { 2 }$

Accepted Answer

By separating variables and integrating, we have:

$$\int \frac{1}{\sqrt{y-1}} dy = \int 8x^7 dx $$

Applying the substitution $u = y - 1$, $du = dy$, we get:

$$2 \sqrt{y-1} = \frac{8}{8} \cdot x^8 + C$$

Solving for $y$, we get:

$$y = \left(\frac{1}{2}x^8 + C\right)^2 + 1$$

Therefore, the answer is (C).

Question 17

Solve the problem.
-The barometric pressure $\mathrm { p }$ at an altitude of $\mathrm { h }$ miles above sea level satisfies the differential equation $\frac { d p } { d h } = - 0.2 p$. If the pressure at sea level is $29.92$ inches of mercury, find the barometric pressure at $29,000 \mathrm { ft }$.
A) 4.99 in.
B) 9.97 in.
C) 89.75 in.
D) 0.09 in.

Accepted Answer

First, we need to convert the altitude from feet to miles.

$29,000 	ext{ ft} = \dfrac{29,000}{5,280} 	ext{ miles} \approx 5.5 	ext{ miles}$

Now we can solve the differential equation:

$\dfrac{dp}{dh} = -0.2p \Rightarrow \dfrac{dp}{p} = -0.2dh$

Integrating both sides:

$\ln|p| = -0.2h + C$

We can find the value of the constant $C$ by using the initial condition that the pressure at sea level is 29.92 inches of mercury:

$\ln|29.92| = -0.2(0) + C \Rightarrow C = \ln|29.92|$

So the equation for the barometric pressure as a function of altitude is:

$\ln|p| = -0.2h + \ln|29.92| = \ln\left(\dfrac{29.92}{e^{0.2h}}ight)$

Now we can use this equation to find the pressure at an altitude of 5.5 miles.

$p = \pm e^{\ln\left(\dfrac{29.92}{e^{0.2(5.5)}}ight)} \approx 9.97 	ext{ in}$

Since pressure cannot be negative, we take the positive solution of the equation, which gives us a pressure of approximately 9.97 inches of mercury at an altitude of 29,000 feet. Therefore, the answer is B.

Question 18

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function.
-$\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =$
A) $\frac { 3 } { 5 }$
B) $\frac { 5 } { 16 }$
C) $\frac { 5 } { 3 }$
D) $\frac { 5 } { 4 }$

Accepted Answer

Given $\sinh x = \frac{3}{4}$, we can find $\cosh x$ using the identity $\cosh^2 x - \sinh^2 x = 1$. Solving for $\cosh x$, we get $\cosh x = \sqrt{\sinh^2 x + 1} = \sqrt{\left(\frac{3}{4}\right)^2 + 1} = \sqrt{\frac{9}{16} + 1} = \sqrt{\frac{25}{16}} = \frac{5}{4}$. The coth function is defined as $\coth x = \frac{\cosh x}{\sinh x}$, so substituting the values we get $\coth x = \frac{5/4}{3/4} = \frac{5}{3}$.

Question 19

Solve the problem.
-The charcoal from a tree killed in a volcanic eruption contained 68.2% 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) 5700 years
B) 2182 years
C) 1512 years
D) 3147 years

Accepted Answer

Using the half-life formula for exponential decay, we can set up the equation: 
0.682 = (1/2)^(t/5700) 
where t is the age of the tree. 
Solving for t, we get: 
t = (ln 0.682) * 5700 / (ln 1/2) 
t ≈ 3147 
Therefore, the tree is approximately 3147 years old.

Question 20

Solve the problem.
-The amount of alcohol in the bloodstream, $\mathrm { A }$, declines at a rate proportional to the amount, that is, $\frac { \mathrm { dA } } { \mathrm { dt } } = - \mathrm { kA }$. If $\mathrm { k } = 0.5$ 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) 2.8 hr
B) 0.3 hr
C) 2.1 hr
D) 1.4 hr

Accepted Answer

We can use the exponential decay formula for this problem:
$$\mathrm{A(t) = A_0 e^{-kt}},$$
where A(t) is the amount of alcohol in the bloodstream at time t, A0 is the initial amount of alcohol, and k is the rate constant.

We are given that k = 0.5, and we can assume that A0 = 0.10%. Thus, we want to solve for t when A(t) = 0.05%:
$$0.05\% = 0.10\% e^{-0.5t}$$
Simplifying, we get:
$$e^{-0.5t} = 0.5$$
$$t = \frac{-\ln 0.5}{0.5} = 1.386 \approx 1.4\mathrm{~hours}$$
Therefore, the answer is D.

Quiz 8: Integrals and Transcendental Functions

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function. - $\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =$

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

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 8 } { 15 } , \operatorname { sech } x =$

Solve the problem. -Suppose that the amount of oil pumped from a well decreases at the continuous rate of 12% per year. When, to the nearest year, will the well's output fall to one-eighth of its present value?

Solve the differential equation. - $\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$

Solve the differential equation. - $\frac { d y } { d x } = 3 \cos x \sec y$

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function. - $\sinh x = - \frac { 4 } { 3 } , \tanh x =$

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 5 } { 12 } , \cosh x =$

Solve the differential equation. - $\frac { d y } { d x } = 6 x \sqrt { 9 - y ^ { 2 } }$

Solve the differential equation. - $\frac { d y } { d x } = 8 x ^ { 7 } \sqrt { y - 1 }$

A value of $\sinh x$ or $\cosh x$ is given. Use the definitions and the identity $\cosh ^ { 2 } x - \sinh ^ { 2 } x = 1$ to find the value of the other indicated hyperbolic function. - $\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =$

Solve the problem. -The charcoal from a tree killed in a volcanic eruption contained 68.2% 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.

Quiz 8: Integrals and Transcendental Functions

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

Solve the differential equation. - dydx=8x7sec⁡y\frac { d y } { d x } = 8 x ^ { 7 } \sec ydxdy​=8x7secy

Solve the problem. -Suppose that the amount of oil pumped from a well decreases at the continuous rate of 12% 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=3x2cos⁡2y\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } ydxdy​=3x2cos2y

Solve the differential equation. - dydx=3cos⁡xsec⁡y\frac { d y } { d x } = 3 \cos x \sec ydxdy​=3cosxsecy

Solve the differential equation. - dydx=6x9−y2\frac { d y } { d x } = 6 x \sqrt { 9 - y ^ { 2 } }dxdy​=6x9−y2​

Solve the differential equation. - dydx=8x7y−1\frac { d y } { d x } = 8 x ^ { 7 } \sqrt { y - 1 }dxdy​=8x7y−1​