Textbook Question
Evaluating hyperbolic functions Evaluate each expression without using a calculator or state that the value does not exist. Simplify answers as much as possible.
f. sinh (2 ln 3)
Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243
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Table of contents
- Ch. 1 - Functions210
- Ch. 2 - Limits371
- Ch. 3 - Derivatives633
- Ch. 4 - Applications of the Derivative412
- Ch. 5 - Integration328
- Ch. 6 - Applications of Integration331
- Ch. 7 - Logarithmic, Exponential Functions, and Hyperbolic Functions177
- Ch. 8 - Integration Techniques394
- Ch. 9 - Differential Equations179
- Ch. 10 - Sequences and Infinite Series339
- Ch. 11 - Power Series218
- Ch.12 - Parametric and Polar Curves215
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
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Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.43d
Briggs 3rd EditionCh. 7 - Logarithmic, Exponential Functions, and Hyperbolic FunctionsProblem 7.3.43dChapter 7, Problem 7.3.43d
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.
d. If the rate constant of an exponential growth function is increased, its doubling time is decreased.
Verified step by step guidance1
Recall the general form of an exponential growth function: \(N(t) = N_0 e^{kt}\), where \(k\) is the rate constant and \(k > 0\) for growth.
Understand that the doubling time \(T\) is the time it takes for the quantity to double, so \(N(T) = 2N_0\).
Set up the equation for doubling time: \(2N_0 = N_0 e^{kT}\), which simplifies to \(2 = e^{kT}\).
Take the natural logarithm of both sides to solve for \(T\): \(\ln(2) = kT\), so \(T = \frac{\ln(2)}{k}\).
Analyze the relationship: since \(\ln(2)\) is constant, if the rate constant \(k\) increases, the denominator increases, making \(T\) smaller. Therefore, increasing the rate constant decreases the doubling time.
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Key Concepts
Here are the essential concepts you must grasp in order to answer the question correctly.
Exponential Growth Function
An exponential growth function models quantities that increase at a rate proportional to their current value, typically expressed as f(t) = f_0 * e^(kt), where k > 0 is the growth rate constant. This function describes processes like population growth or compound interest.
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Exponential Growth & Decay
Rate Constant and Its Effect
The rate constant k in an exponential growth function determines how quickly the quantity grows. A larger k means the function grows faster, causing the value to increase more rapidly over time compared to a smaller k.
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Doubling Time in Exponential Growth
Doubling time is the time required for an exponentially growing quantity to double in size. It is calculated as T = ln(2)/k, showing that as the rate constant k increases, the doubling time decreases, meaning the quantity doubles faster.
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Related Practice
Textbook Question
Chemotherapy In an experimental study at Dartmouth College, mice with tumors were treated with the chemotherapeutic drug Cisplatin. Before treatment, the tumors consisted entirely of clonogenic cells that divide rapidly, causing the tumors to double in size every 2.9 days. Immediately after treatment, 99% of the cells in the tumor became quiescent cells which do not divide and lose 50% of their volume every 5.7 days. For a particular mouse, assume the tumor size is 0.5 cm³ at the time of treatment.
d. Plot a graph of V(t) for 0 ≤ t ≤ 15. What happens to the size of the tumor, assuming there are no follow-up treatments with Cisplatin?
Textbook Question
Evaluating hyperbolic functions Evaluate each expression without using a calculator or state that the value does not exist. Simplify answers as much as possible.
d. sech (sinh 0)
Textbook Question
Energy consumption On the first day of the year (t=0), a city uses electricity at a rate of 2000 MW. That rate is projected to increase at a rate of 1.3% per year.
c. Find a function that gives the total energy used (in MW-yr) between t=0 and any future time t>0.
Textbook Question
Terminal velocity Refer to Exercises 95 and 96.
d. How tall must a cliff be so that the BASE jumper (m = 75 kg and k = 0.2) reaches 95% of terminal velocity? Assume the jumper needs at least 300 m at the end of free fall to deploy the chute and land safely.
Textbook Question
Explain why or why not Determine whether the following statements are true and give an explanation or counterexample. Assume x > 0 and y > 0.
e. The area under the curve y = 1/x and the x-axis on the interval [1, e] is 1.