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.
-$\sinh x = - \frac { 12 } { 5 } , \operatorname { csch } x =$

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.

Accepted Answer

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

Question 3

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

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.
-$\sinh x = \frac { 8 } { 15 } , \operatorname { sech } x =$

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?

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 $\mathrm { y } = \mathrm { y } _ { 0 } \mathrm { e } ^ { \mathrm { kt } }$.

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?

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.
-$$\frac { d y } { d x } = 3 x ^ { 2 } \cos ^ { 2 } y$$

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.
-$$\frac { d y } { d x } = 3 \cos x \sec y$$

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?

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.
-$\sinh x = - \frac { 4 } { 3 } , \tanh x =$

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.
-$\sinh x = \frac { 5 } { 12 } , \cosh x =$

Accepted Answer

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

Question 13

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

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 $185 ^ { \circ } \mathrm { F }$.

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.

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

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 $29,000 \mathrm { ft }$.

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.
-$\sinh x = \frac { 3 } { 4 } , \operatorname { coth } x =$

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.

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.

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​