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Derivatives as Rates of Change III

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Recall

The derivative of a function at a chosen input value, when it exists, is the slope of the tangent line to the graph of the function at that point. It is the instantaneous rate of change, the ratio of the instantaneous change in the dependent variable to that of the independent variable.

Definition. A function f(x) is differentiable at a point “a” of its domain, if its domain contains an open interval containing “a”, and the limit $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$ exists, f’(a) = L = $\lim _{h \to 0}{\frac {f(a+h)-f(a)}{h}}$. More formally, for every positive real number ε, there exists a positive real number δ, such that for every h satisfying 0 < |h| < δ, then |L-$\frac {f(a+h)-f(a)}{h}$|< ε.

Martha loves Calculus

Martha loves Calculus

 

  1. Power Rule: $\frac{d}{dx}(x^n) = nx^{n-1}$.
  2. Sum Rule: $\frac{d}{dx}(f(x) + g(x)) = \frac{d}{dx}(f(x)) + \frac{d}{dx}(g(x))$
  3. Product Rule: $\frac{d}{dx}(f(x) \cdot g(x)) = f’(x)g(x) + f(x)g’(x)$.
  4. Quotient Rule: $\frac{d}{dx}\left(\frac{f(x)}{g(x)}\right) = \frac{f’(x)g(x) - f(x)g’(x)}{(g(x))^2}$
  5. Chain Rule: $\frac{d}{dx}(f(g(x))) = f’(g(x)) \cdot g’(x)$
  6. $\frac{d}{dx}(e^x) = e^x, \frac{d}{dx}(\ln(x)) = \frac{1}{x}, \frac{d}{dx}(\sin(x)) = \cos(x), \frac{d}{dx}(\cos(x)) = -\sin(x), \frac{d}{dx}(\tan(x)) = \sec^2(x), \frac{d}{dx}(\arcsin(x)) = \frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arccos(x)) = -\frac{1}{\sqrt{1 - x^2}}, \frac{d}{dx}(\arctan(x)) = \frac{1}{1 + x^2}.$

The critical points of a function f are the x-values, within the domain (D) of f for which f’(x) = 0 or where f’ is undefined. Notice that the sign of f’ must stay constant between two consecutive critical points. If the derivative of a function changes sign around a critical point, the function is said to have a local or relative extremum (maximum or minimum) at that point. If f’ changes sign from positive (increasing function) to negative (decreasing function), the function has a local or relative maximum at that critical point. Similarly, if f’ changes sign from negative to positive, the function has a local or relative minimum.

Interpretation

f’(x) is the slope of the line tangent to the graph of f at that particular point (x, f(x)). f’(x) is also the rate of change of the function at x. The average rate of change is the process of calculating the rate at which the output (y-values) changes compared to its input (x-values). This can be visualized as the slope of a secant line passing between two points on a function. In differential calculus, the focus shifts to the instantaneous rate of change, which is found using the derivative of a function.

For example, Growth Rate = $\frac{Births-Deaths}{years}$. The rate of change in population is the derivative of the population function with respect to time, $\frac{dP}{dt}$.

How to solve rates of change problems

You should follow these general steps:

  1. Understand the problem, draw a diagram, identify what quantity is changing and with respect to what other quantity.
  2. Label all quantities and their rates of change.
  3. Relate all quantities in a single equation, that is, use the given information and the problem constraints to set up an equation that relates the quantities involved in the same equation.
  4. Differentiate the equation: Calculate the derivative of the equation with respect to the relevant variable to obtain the rate of change.
  5. Solve for the unknown: Plug in the known values and solve for the unknown rate of change to answer the question.

Related rates

 

The diagram is shown in Figure 1. By the Pythagorean theorem, x2 + y2 = 102.

Let’s differentiate by t the previous formula, $2x·\frac{dx}{dt}+2y·\frac{dy}{dt} = 0$ (🚀). Our goal is to calculate $\frac{dy}{dt}$ when x = 6 (⇒ y = $\sqrt{10^2-6^2} = \sqrt{100-36} = \sqrt{64} = 8$), and we also know that $\frac{dx}{dt}=1$, so let’s plug in all of these values in (🚀) $2x·\frac{dx}{dt}+2y·\frac{dy}{dt} = 0 ↭ 2·6·1 + 2·8·\frac{dy}{dt} = 0 ⇒ \frac{dy}{dt} = \frac{-12}{16} = \frac{-3}{4}.$ The top of the ladder is sliding down at a rate of -0.75m/s.

 

The diagram is shown in Figure 2. By the Pythagorean theorem, A2 +B2 = C2. We differentiate by t, $2·A·\frac{dA}{dt}+2·B·\frac{dB}{dt} = 2·C·\frac{dC}{dt}$ [🚀]

We are asked $\frac{dC}{dt}$ when car A is 0.3 meters and car B is 0.4 from the intersection. A = 0.3, B = 0.4 ⇒ $0.3^2+0.4^2 = C^2 ⇒ C = \sqrt{0.25} = 0.5$. We are also told that $\frac{dA}{dt} = -50, \frac{dB}{dt} = -60.$

Let’s plug in all these values in [🚀] ⇒ $2·A·\frac{dA}{dt}+2·B·\frac{dB}{dt} = 2·C·\frac{dC}{dt} ⇒ 2·0.3·(-50) + 2·0.4·(-60) = 2·0.5\frac{dC}{dt} ⇒ \frac{dC}{dt} = -78m/h$, that is, the cars are approaching each other at a rate of 78m/h.

$\frac{dV}{dt}=12.8·π$. Goal: r when $\frac{dr}{dt} = 0.2$?

Recall that the volume of a sphere is V = $\frac{4}{3}πr^3 ⇒ \frac{dV}{dr} = 4·π·r^2 ⇒ \frac{dr}{dV} = \frac{1}{4·π·r^2}.$

$\frac{dr}{dt} = \frac{dr}{dV}·\frac{dV}{dt} ⇒ 0.2 = \frac{dr}{dV}·\frac{dV}{dt} = \frac{1}{4·π·r^2}·12.8 ⇒ r^2 = \frac{12.8·π}{0.2·4·π} = \frac{12.8}{0.8} = 16 ⇒ r = ±4$ ⇒[A radius is obviously positive] r = 4m.

The diagram is shown in Figure E. By the Pythagorean theorem, x2 + y2 = z2 ⇒[We differentiate by t] $2x\frac{dx}{dt} +2y\frac{dy}{dt}= 2z\frac{dz}{dt}$

Our goal is $\frac{dx}{dt}$ when y = 0.6, x = 0.8 (x2 + y2 = z2 ↭ $z = \sqrt{0.6^2+0.8^2}$ = 1), $\frac{dz}{dt} = 20, \frac{dy}{dt} = -60$.

Let’s plug in these values into the previous formula, $2x\frac{dx}{dt} +2y\frac{dy}{dt}= 2z\frac{dz}{dt} ⇒ 2·0.8·\frac{dx}{dt} +2·0.6·(-60) = 2·1·20 ⇒ 1.6·\frac{dx}{dt} = 40+72 ⇒ \frac{dx}{dt} = \frac{40+72}{1.6} = 70mph.$

a = 5,000 meters. $\frac{db}{dt}\bigg|_{10,000} = 200m/s$

a2 + b2 = c2. In that particular moment, $5,000^2+10,000^2 = c^2⇒ c = \sqrt{5,000^2+10,000^2} = \sqrt{5,000^2+5,000^2·2^2} = 5,000\sqrt{1+2^2} = 5,000·\sqrt{5}$

a2 + b2 = c2 ⇒ $\frac{d(a^2+b^2)}{dt} = \frac{dc^2}{dt} ⇒ 2a·\frac{da}{dt}+2b·\frac{db}{dt} = 2c\frac{dc}{dt} ⇒ 2·5,000·0+2·10,000·200 = 2·5,000·\sqrt{5}·\frac{dc}{dt} ⇒ 4,000,000 = 10,000\sqrt{5}·\frac{dc}{dt} ⇒ \frac{dc}{dt}=\frac{4,000,000}{10,000·\sqrt{5}} = \frac{400}{\sqrt{5}}=\frac{400·\sqrt{5}}{\sqrt{5}·\sqrt{5}} = \frac{400·\sqrt{5}}{5} = 80·\sqrt{5}m/s.$

$\frac{dV}{dt}=150, \frac{dh}{dt}\bigg|_{h=1.4}?$

V = $3·(2 + \sqrt{h})^6-192 ⇒ \frac{dV}{dt} = 150 = 3·6·(2 + \sqrt{h})^5\frac{1}{2\sqrt{h}}\frac{dh}{dt}⇒\frac{dh}{dt} = \frac{150·\sqrt{h}}{9·(2 + \sqrt{h})^5}$

$\frac{dh}{dt}\bigg|_{h=1.4} = \frac{150·\sqrt{1.4}}{9·(2 + \sqrt{1.4})^5}$ ≈ 0.06 m/s.

We know that 32 + b2 = c2, $\frac{db}{dt}=480$

480 miles ———- 1h = 60*60 seconds
b miles ———— 30 seconds

b = $\frac{480·30}{60·60} = 4 miles$ ⇒[32 + b2 = c2] c = 5.

The diagram is shown in Figure 2.b.  

32 + b2 = c2 ⇒ 2b$\frac{db}{dt} = 2c\frac{dc}{dt}$. Now, we can insert the values in this equation, $2·4·480 = 2·5·\frac{dc}{dt}⇒~ \frac{dc}{dt}=384mph$∎

Bibliography

This content is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License.
  1. NPTEL-NOC IITM, Introduction to Galois Theory.
  2. Algebra, Second Edition, by Michael Artin.
  3. LibreTexts, Calculus. Abstract and Geometric Algebra, Abstract Algebra: Theory and Applications (Judson).
  4. Field and Galois Theory, by Patrick Morandi. Springer.
  5. Michael Penn, and MathMajor.
  6. Contemporary Abstract Algebra, Joseph, A. Gallian.
  7. YouTube’s Andrew Misseldine: Calculus. College Algebra and Abstract Algebra.
  8. MIT OpenCourseWare 18.01 Single Variable Calculus, Fall 2007 and 18.02 Multivariable Calculus, Fall 2007.
  9. Calculus Early Transcendentals: Differential & Multi-Variable Calculus for Social Sciences.
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