EMath 1101 Past Paper PDF

## EMath 1101 Dec 06, 2024 ### Bases Other than e The base of the natural exponential function is e. This “natural” base can be used to assign a meaning to a general base *a*. ### Definition of Exponential Function to Base *a* If *a* is a positive real number (*a* ≠ 1) and *x* is any real numb...

## EMath 1101 Dec 06, 2024 ### Bases Other than e The base of the natural exponential function is e. This “natural” base can be used to assign a meaning to a general base *a*. ### Definition of Exponential Function to Base *a* If *a* is a positive real number (*a* ≠ 1) and *x* is any real number, then the exponential function to the base *a* is denoted by *a^x* and is defined by *a^x* = *e^{(ln a) x}*. If *a* = 1, then *y* = 1^*x* = 1 is a constant function. These functions obey the usual laws of exponents. For instance, here are some familiar properties. - *a⁰* = 1 - *a^x* *a^y* = *a^{x + y}* - *a^x* / *a^y* = *a^{x - y}* - (*a^x*)^*y* = *a^xy* ### Example: Radioactive Half-Life Model The half-life of carbon-14 is about 5715 years. A sample contains 1 gram of carbon-14. How much will be present in 10,000 years? **Solution:** Let *t* = 0 represent the present time and let *y* represent the amount (in grams) of carbon-14 in the sample. Using a base of *½*, you can model *y* by the equation: *y* = *(½)^{(t/ 5715)}* Notice that when *t* = 5715, the amount is reduced to half of the original amount. *y* = *(½)^{(5715/ 5715)}* = *½* gram When *t* = 11, 430, the amount is reduced to a quarter of the original amount, and so on. To find the amount of carbon-14 after 10,000 years, substitute 10,000 for *t*. *y* = *(½)^{(10,000/ 5715)}* ≈ 0.30 gram The graph of *y* is shown in the attached image. ### Logarithmic Functions to Bases Other than *e* Logarithmic functions to bases other than *e* can be defined in much the same way as exponential functions to other bases are defined. ### Definition of Logarithmic Function to Base *a* If *a* is a positive real number (*a* ≠ 1) and *x* is any positive real number, then the logarithmic function to the base *a* is denoted by *log_ax* and is defined as: *log_ax* = *1 / ln(a) x*. Logarithmic functions to the base *a* have properties similar to those of the natural logarithmic function given in Theorem 5.2. (Assume x and y are positive numbers and *n* is rational.) - *log_a1* = 0 - *log_axy* = *log_ax* + *log_ay* - *log_axⁿ* = *n log_ax* - *log_a(x/y)* = *log_ax* - *log_ay* ### Properties of Inverse Functions From the definitions of the exponential and logarithmic functions to the base *a*, it follows that *f(x)* = *a^x* and *g(x)* = *log_ax* are inverse functions of each other. - *y* = *a^x* if, and only if, *x* = *log_ay* - *a^log_ax* = *x* for *x* > 0 - *log_aa^x* = *x* for all *x* ### Example: Bases Other than *e* Solve for *x* in each equation: a. *3^x* = *1/81* **Solution:** To solve this equation, you can apply the logarithmic function to the base 3 to each side of the equation. *log₃3^x* = *log₃1/81* *x* = *log₃3^-4* *x* = *-4* b. *log₂x* = *-4* **Solution:** To solve this equation, you can apply the exponential function to the base 2 to each side of the equation. *2^log₂x* = *2^-4* *x* = *1/2⁴* *x* = *1/16* ### Derivatives for Bases Other than *e* Let *a* be a positive real number (*a* ≠ 1), and let *u* be a differentiable function of *x*. - *d/dx*[*a^x*] = *(ln a)a^x* - *d/dx*[*a^u*] = *(ln a)a^u du/dx* - *d/dx*[*log_ax*] = *1 / (ln a)x* - *d/dx*[*log_au*] = *1 / (ln a)u du/dx* ### Example: Differentiating Functions to Other Bases Find the derivative of each function: a. *y* = *2^x* **Solution:** *y'* = *d/dx*[*2^x*] = *(ln 2)2^x* b. *y* = *2^3x* **Solution:** *y'* = *d/dx*[*2^3x*] = *(ln 2)2^3x(3)* = *(3 ln 2)2^3x* Try writing *2^3x* as *8^x* and differentiating to see that you obtain the same result. c. *y* = *log₁₀cos x* **Solution:** *y'* = *d/dx*[*log₁₀cos x*] = *-sin x * 1/(ln 10)cos x* = *-1/ln 10 tan x*. ### The Power Rule for Real Exponents Let *n* be any real number and let *u* be a differentiable function of *x*. - *d/dx*[*xⁿ*] = *nx^{n - 1}* - *d/dx*[*uⁿ*] = *nu^n-1 du/dx* ### Example: Comparing Variables and Constants a. *d/dx*[e] = 0* **Constant Rule** b. *d/dx[e^x] = e^x* **Exponential Rule** c. *d/dx[x^x] = xe^x-1* **Power Rule** d. *y* = *x^x **Logarithmic Differentiation** *ln y* = *ln x^x* *ln y* = *x ln x* *y'/y* = *(1/x)* + (ln *x*)(1) = *1* + *ln x* *y'* = *y*(1 + *ln x*) = *x^x*(1 + *ln x*) ### Definitions of Inverse Trigonometric Functions | Function | Domain | Range | |---|---|---| | *y* = *arcsin x* iff *sin y* = *x* | -1 ≤ *x* ≤ 1 | -π / 2 ≤ *y* ≤ π / 2 | | *y* = *arccos x* iff *cos y* = *x* | -1 ≤ *x* ≤ 1 | 0 ≤ *y* ≤ π | | *y* = *arctan x* iff *tan y* = *x* | -∞ < *x* < ∞ | -π / 2 < *y* < π / 2 | | *y* = *arccot x* iff *cot y* = *x* | -∞ < *x* < ∞ | 0 < *y* < π | | *y* = *arcsec x* iff *sec y* = *x* | |*x|* ≥ 1 | 0 ≤ *y* ≤ π, *y* ≠ π / 2 | | *y* = *arccsc x* iff *csc y* = *x* | |*x|* ≥ 1 | -π / 2 ≤ *y* ≤ π / 2, *y* ≠ 0 | ### Example: Evaluating Inverse Trigonometric Functions Evaluate each function: a. *arcsin(-½)* **Solution:** By definition, *y* = *arcsin(-½)* implies that *sin y* = -½. In the interval [-π / 2, π / 2], the correct value of *y* is -π / 6. *arcsin(-½)* = *-π / 6* b. *arccos 0* **Solution:** By definition, *y* = *arccos 0* implies that *cos y* = 0. In the interval [0, π], you have *y* = π / 2. *arccos 0* = *π / 2* c. *arctan √3* **Solution:** By definition, *y* = *arctan √3* implies that *tan y* = √3. In the interval (-π / 2, π / 2), you have *y* = π / 3. *arctan √3* = *π / 3* d. *arcsin(0.3)* **Solution:** Using a calculator set in radian mode produces *arcsin(0.3)* ≈ 0.305. ### Properties of Inverse Trigonometric Functions - If -1 ≤ *x* ≤ 1 and -π / 2 ≤ *y* ≤ π / 2, then: >*sin(arcsin x)* = *x* >*arcsin(sin y)* = *y* - If -π / 2 < *y* < π / 2, then: >*tan(arctan x)* = *x* >*arctan(tan y)* = *y* - If |*x|* ≥ 1 and 0 ≤ *y* < π / 2 or π / 2 < *y* ≤ π, then: >*sec(arcsec x)* = *x* >*arcsec(sec y)* = *y* Similar properties hold for the other inverse trigonometric functions. ### Example: Solving an Equation *arctan(2x - 3)* = *π / 4* **Solution:** *tan[arctan(2x - 3)]* = *tan π / 4* *2x - 3* = *1* *x* = *2* Some problems in calculus require that you evaluate expressions such as *cos(arcsin x)*, as shown in the example below. ### Example: Using Right Triangles a. Given *y* = *arcsin x*, where *0 < y < π / 2*, find *cos y*. b. Given *y* = *arcsec(√5 / 2)*, find *tan y*. **Solution:** a. Because *y* = *arcsin x*, you know that *sin y* = *x*. This relationship between *x* and *y* can be represented by a right triangle, as shown in the attached image. *cos y* = *cos(arcsin x)* = *adj. / hyp.* = √(1 - *x²*) / 1 (This result is also valid for -π / 2 < * y* < 0.) b. Use the right triangle shown in the attached image. *tan y* = *tan arcsec (√5 / 2)* = *opp. / adj.* = 1 / 2 ### Derivatives of Inverse Trigonometric Functions Let *u* be a differentiable function of *x*. - *d/dx*[arcsin u]* = *u' / √(1 - u²)* - *d/dx*[arccos u]* = *-u' / √(1 - u²)* - *d/dx*[arctan u]* = *u' / (1 + u²)* - *d/dx*[arccot u]* = *-u' / (1 + u²)* - *d/dx*[arcsec u]* = *u' / |u|√(u² - 1)* - *d/dx*[arccsc u]* = *-u' / |u|√(u² - 1)* ### Example: Differentiating Inverse Trigonometric Functions a. *d/dx*[arcsin(2x)]* = *2 / √(1 - (2x)²)* = *2 / √(1 - 4x²)* b. *d/dx*[arctan(3x)]* = *3 / (1 + (3x)²)* = *3 / (1 + 9x²)* c. *d/dx*[arcsin√x]* = *(1/2)x^-1/2 / √(1 - x)* = *1 / 2√x√(1 - x)* = *1 / 2√(x - x²)* d. *d/dx*[arcsec e^2x]* = *2e^2x / e^2x√((2x)² - 1)* = *2 / √(e^4x - 1)* The absolute value sign is not necessary because e^2x > 0. ### Example: A Derivative That Can Be Simplified *y* = *arcsin x + x√(1 - x²)* **Solution:** *y'* = *1 / √(1 - x²)* + *x(2)(-2x)(1 - x²)^-1/2+ √(1 - *x²)* = *1 / √(1 - x²)* - *x² / √(1 - x²)* + √(1 - *x²)* = *2√(1 - x²)* ### Example: Analyzing an Inverse Trigonometric Graph Analyze the graph of *y* = *(arctan x)²*. **Solution:** From the derivative: *y’* = 2(*arctan x*)(*1* + *x²*) = *2arctan x / (1 + x²)* you can see that the only critical number is *x* = 0. By the First Derivative Test, this value corresponds to a relative minimum. From the second derivative: *y”* = *(1 + x²)²* - *(2 arctan x)(2x) / (1 + x²)²* = *2(1 - 2x arctan x) / (1 + *x²)²* it follows that points of inflection occur when *2x arctan x* = *1*. Using Newton’s Method, these points occur when *x* ≈ ±0.765. Finally, because: lim *x*→∞ (*arctan x*)²* = *π² / 4* it follows that the graph has a horizontal asymptote at *y* = *π² / 4*. The graph is shown in the attached image. ### Basic Differentiation Rules for Elementary Functions | Rule | Formula | |---|---| | 1 | *d/dx*[c*u*] = c*u’* | | 2 | *d/dx*[u ± v]* = *u’ ± v’* | | 3 | *d/dx*[u*v*] = *u*v’ + *v*u’* | | 4 | *d/dx*[u/v]* = (*v*u’ - *u*v’)/ *v²* | | 5 | *d/dx*[c]* = 0 | | 6 | *d/dx*[uⁿ]* = *n*u^n-1*u’* | | 7 | *d/dx*[x]* = 1 | | 8 | *d/dx*[|u|]* = *u’ / |u|*, *u* ≠ 0 | | 9 | *d/dx*[ln u]* = *u’ / u* | | 10 | *d/dx*[e^u]* = *e^u* u’* | | 11 | *d/dx*[log_au]* = *(u’ / ln a)u* | | 12 | *d/dx*[a^u]* = *(ln a)a^uu’* | | 13 | *d/dx*[sin u]* = *(cos u)u’* | | 14 | *d/dx*[cos u]* = *-(sin u)u’* | | 15 | *d/dx*[tan u]* = *(sec² u)u’* | | 16 | *d/dx*[cot u]* = *-(csc² u)u’* | | 17 | *d/dx*[sec u]* = *(sec u tan u)u’* | | 18 | *d/dx*[csc u]* = *-(csc u cot u)u’* | | 19 | *d/dx*[arcsin u]* = *u’ / √(1 - u²)* | | 20 | *d/dx*[arccos u]* = *-u’ / √(1 - u²)* | | 21 | *d/dx*[arctan u]* = *u’ / (1 + u²)* | | 22 | *d/dx*[arccot u]* = *-u’ / (1 + u²)* | | 23 | *d/dx*[arcsec u]* = *u’ / |u|√(u² - 1)* | | 24 | *d/dx*[arccsc u]* = *-u’ / |u|√(u² - 1)* | ### Hyperbolic Identities - *cosh²x* - *sinh²x* = 1 - *tanh²x* + *sech²x* = 1 - *coth²x* - *csch²x* = 1 - *sinh(x + y)* = *sinh x cosh y* + *cosh x sinh y* - *sinh(x - y)* = *sinh x cosh y* - *cosh x sinh y* - *cosh(x + y)* = *cosh x cosh y* + *sinh x sinh y* - *cosh(x - y)* = *cosh x cosh y* - *sinh x sinh y* - *sinh²x* = (-1 + *cosh 2x*) / 2 - *cosh²x* = (1 + *cosh 2x*) / 2 - *sinh 2x* = 2 *sinh x cosh x* - *cosh 2x* = *cosh²x* + *sinh²x* ### Derivatives and Integrals of Hyperbolic Functions Let *u* be a differentiable function of *x*. | Rule | Formula | |---|---| | *d/dx*[sinh u]* = *(cosh u)u’* | ∫cosh u du = sinh u + C | | *d/dx*[cosh u]* = *(sinh u)u’* | ∫sinh u du = cosh u + C | | *d/dx*[tanh u]* = *(sech² u)u’* | ∫sech² u du = tanh u + C | | *d/dx*[coth u]* = *-(csch² u)u’* | ∫csch² u du = -coth u + C | | *d/dx*[sech u]* = *-(sech u tanh u)u’* | ∫sech u tanh u du = -sech u + C | | *d/dx*[csch u]* = *-(csch u coth u)u’* | ∫csch u coth u du = -csch u + C | ### Example: Differentiation of Hyperbolic Functions a. *d/dx*[sinh(x² - 3)]* = *2x cosh(x² - 3)* b. *d/dx*[ln(cosh x)]* = *sinh x / cosh x* = *tanh x* c. *d/dx*[x sinh x - cosh x]* = *x cosh x + sinh x - sinh x* = *x cosh x* ### Example: Finding Relative Extrema Find the relative extrema of *f(x)* = *(x - 1) cosh x - sinh x*. **Solution:** Begin by setting the first derivative of *f* equal to 0. *f’(x)* = *(x - 1) sinh x + cosh x cosh x* = 0 *(x - 1) sinh x + cosh²x* = 0 *(x - 1) sinh x* = 0 So, the critical numbers are *x* = 1 and *x* = 0. Using the Second Derivative Test, you can verify that the point (0, -1) yields a relative maximum and the point (1, -sinh 1) yields a relative minimum, as shown in the attached image. Try using a graphing utility to confirm this result. If your graphing utility does not have hyperbolic functions, you can use exponential functions, as follows. *f(x)* = *(x - 1)(½)(e^x + e^-x) - (e^x - e^-x)* = *(xe^x + xe^-x - e^x - e^-x - e^x + e^-x)* = *(xe^x + xe^-x - 2e^x)* When a uniform flexible cable, such as a telephone wire, is suspended from two points, it takes the shape of a catenary, as discussed in the example below. ### Example: The Catenary The shape of a hanging cable, like a telephone wire, is described by the equation *y* = *a cosh(x/a)*. The constant *a* represents the tension in the cable. This curve is called a catenary.