Week 1 Activity

Section 1.3 Week 1 Activity

Subsection 1.3.1 Hyperbolic Functions

Activity 1.3.1.

Calculate these limits.

\begin{equation*} \lim_{x \rightarrow 0} \frac{\sin x \sinh x}{x^2} \end{equation*}
\begin{equation*} \lim_{x \rightarrow 0} x^3 \coth x \end{equation*}
\begin{equation*} \lim_{x \rightarrow \infty} \frac{\sinh x}{\cosh x} \end{equation*}

Solution.

\begin{equation*} \lim_{x \rightarrow 0} \frac{\sin x \sinh x}{x^2} = \lim_{x \rightarrow 0} \frac{\sin x}{x} \frac{\sinh x}{x} = \lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{\sinh x}{x} = 1 \cdot 1= 1\\ \end{equation*}

This limit can be split up into two limits. For both \(\sin x\) and \(\sinh x\text{,}\) there are known limits for the ratio with \(x\) near zero.
\begin{equation*} \lim_{x \rightarrow 0} x^3 \coth x = \lim_{x \rightarrow 0} \frac{x}{\sinh x} x^2 \cosh x = = \left( \lim_{x \rightarrow 0} \frac{x}{\sinh x} \right) \left( \lim_{x \rightarrow 0} x^2 \right) \left( \lim_{x \rightarrow 0} \cosh x \right)= 1 \cdot 0 \cdot 1 = 0 \end{equation*}

By writing hyperbolic cotangent in terms of hyperbolic sine and cosine, this limit can also be split up. Then it uses the known limit for \(\frac{\sinh x}{x}\) and simply evaluate the other two pieces.
\begin{equation*} \lim_{x \rightarrow \infty} \frac{\sinh x}{\cosh x} = \lim_{x \rightarrow \infty} \frac{e^x - e^{-x}}{e^x + e^{-x}} = 1 \end{equation*}

For this limit, I can write the hyperbolics in terms of their exponential defintions and then use asymptotic analysis. I see that the numerator and denominator have the same order and both have leading coefficients \(1\text{.}\) (For limits at infinity, I will almost always rely on exponential definitions of the hyperbolics, either directly on indirectly.)

Activity 1.3.2.

Calculate these derivatives.

\begin{equation*} \frac{d}{dx} e^{\cosh x} \end{equation*}
\begin{equation*} \frac{d}{dx} \cosh (1 + x^2) \end{equation*}

Solution.

\begin{align*} \frac{d}{dx} e^{\cosh x} \amp = \frac{d}{du} e^u \Big|_{u = \cosh x} \frac{d}{dx} \cosh x\\ \amp = e^u \Big|_{u = \cosh x} \sinh x = e^{\cosh x} \sinh x \end{align*}

This is the chain rule with \(u = \cosh x\text{.}\)
\begin{align*} \amp \frac{d}{dx} \cosh (1 + x^2) = \frac{d}{du} \cosh u \Big|_{u = 1 + x^2} \frac{d}{dx} (1+x^2) \\ \amp = \sinh u \Big|_{u = 1 + x^2} (2x) = \sinh (1+x^2) (2x) = 2x \sinh (1+x^2) \end{align*}

This is the chain rule with \(u = 1+x^2\text{.}\)

Activity 1.3.3.

Calculate these integrals.

\begin{equation*} \int \sinh (x^6) x^5 dx \end{equation*}
\begin{equation*} \int \cosh^3 x \sinh 2x dx \end{equation*}

Solution.

\begin{equation*} \int \sinh (x^6) x^5 dx = \frac{1}{6} \int \sinh u du = \frac{1}{6} \cosh u + c = \frac{1}{6} \cosh (x^6) + c \end{equation*}

In the first step, I used substitution with \(u = x^6\) and \(du = 6x^5dx \text{,}\) so \(\frac{1}{6} du = x^5 dx\text{.}\) After the replacements are done, the antiderivative of \(\sinh u\) is \(\cosh u\text{.}\) Then I reverse the substitution and add a constant of integration.
\begin{equation*} \int \cosh^3 x \sinh 2x dx = \int \cosh^3 x 2 \sinh x \cosh x dx \end{equation*}

The \(\sinh 2x\) term makes the integral difficult to approach. Happily, there is a double-angle identity for this. I replaced \(\sinh 2x\) with \(2 \sinh x \cosh x\) from the table of identities. Then I group the \(\cosh x\) terms together using exponents.

\begin{equation*} = 2 \int \cosh^4 x \sinh x dx \end{equation*}

This can now be solved by substitution with \(u = \cosh x\) and \(du = \sinh x dx\text{.}\) After the replacements, I get a power rule integral.

\begin{equation*} = 2 \int u^4 du = 2 \frac{u^5}{5} + c = \frac{2 \cosh^5 x}{5} + c \end{equation*}

After doing the antiderivative, I reversed the substitution.

Subsection 1.3.2 Implicit Derivatives

Activity 1.3.4.

Calculate the implicit derivative for the following locus. Sketch the locus and observe the expected slopes; compare those with the slopes given by the implicit derivatives.

\begin{equation*} x^2 + y^2 = 100 \end{equation*}

Solution.

First I differentiate both sides of the equation in the variable \(x\text{.}\) I am implicitly assuming that \(y\) is a function of \(x\text{,}\) so when I differentiate an expression in \(y\text{,}\) the chain rules applies and I multiply by the derivative of the inside, which is the unknown \(\frac{dy}{dx}\text{.}\)

\begin{align*} \frac{d}{dx} x^2 + y^2 \amp = \frac{d}{dx} 100 \\ 2x + 2y \frac{dy}{dx} \amp = 0 \\ \frac{dy}{dx} \amp = -\frac{x}{y} \end{align*}

By solving for \(\frac{dy}{dx}\text{,}\) I get an expression in both \(x\) and \(y\) that describes the derivative. I can evalute this at a few points.

\begin{gather*} \frac{dy}{dx} (0,10) = -\frac{0}{10} = 0 \\ \frac{dy}{dx} \left( \sqrt{50}, \sqrt{50} \right) = - \frac{\sqrt{50}}{\sqrt{50}} = -1 \\ \frac{dy}{dx} \left( \sqrt{30}, -\sqrt{70} \right) = -\frac{\sqrt{30}}{-\sqrt{70}} = \sqrt{\frac{3}{7}} \end{gather*}

Also, the derivative at \((-10,0)\) is undefined due to division by zero. Possibly there is a vertical tangent line at this point.

Figure 1.3.1. Implicit Derivatives to \(x^2 + y^2 = 100\)

Figure 1.3.1 shows the graph of the locus. Looking at the three points where I evaluted the derivaitve, I can see that the tangent lines have slopes which match my calculations.

Activity 1.3.5.

Calculate the implicit derivative for the following locus. Sketch the locus and observe the expected slopes; compare those with the slopes given by the implicit derivatives.

\begin{equation*} \frac{x^2}{4} + \frac{y^2}{64} = 1 \end{equation*}

Solution.

\begin{gather*} \frac{d}{dx} \frac{x^2}{4} + \frac{y^2}{64} = \frac{d}{dx} 1 \\ \frac{x}{2} + \frac{y}{32} \frac{dy}{dx} = 0\\ \frac{dy}{dx} = - \frac{16x}{y} \end{gather*}

By solving for \(\frac{dy}{dx}\text{,}\) I get an expression in both \(x\) and \(y\) that describes the derivative. I can evalute this at a few points.

\begin{gather*} \frac{dy}{dx} (0,8) = \frac{-0}{8} = 0 \\ \frac{dy}{dx} (\sqrt{2},\sqrt{32}) = \frac{-16\sqrt{2}}{\sqrt{32}} = \frac{-16}{\sqrt{16}} = -4 \\ \frac{dy}{dx} (1,-\sqrt{48}) = \frac{-16}{-\sqrt{48}} = \frac{4}{\sqrt{3}} \end{gather*}

Also, at \((-2,0)\text{,}\) the derivative is undefined due to division by zero. Possibly there is vertical tangent here.

Figure 1.3.2. Implicit Derivatives to \(\frac{x^2}{4} + \frac{y^2}{64} = 1\)

Figure 1.3.2 shows the graph of the locus. Looking at the three points where I evaluted the derivaitve, I can see that the tangent lines have slopes which match my calculations.

Activity 1.3.6.

Calculate the implicit derivative for the following locus. Sketch the locus and observe the expected slopes; compare those with the slopes given by the implicit derivatives.

\begin{equation*} x^2 - y^2 = 4 \end{equation*}

Solution.

\begin{align*} \frac{d}{dx} x^2 - y^2 \amp = \frac{d}{dx} 4 \\ 2x - 2y \frac{dy}{dx} \amp = 0 \\ \frac{dy}{dx} = \frac{x}{y} \end{align*}

By solving for \(\frac{dy}{dx}\text{,}\) I get an expression in both \(x\) and \(y\) that describes the derivative. I can evalute this at a few points.

\begin{gather*} \frac{dy}{dx} (6,-\sqrt{32}) = \frac{-6}{\sqrt{32}} = - \frac{6}{\sqrt{32}} \\ \frac{dy}{dx} (-6,-\sqrt{32}) = \frac{-6}{-\sqrt{32}} = \frac{6}{\sqrt{32}} \end{gather*}

In addition, the slopes at \((-2,0)\) and \((2,0)\) are undefined due to division by zero. Possibly there are vertical tangents here.

Figure 1.3.3. Implicit Derivatives to \(x^2 - y^2 = 4\)

Figure 1.3.3 shows the graph of the locus. Looking at the three points where I evaluted the derivaitve, I can see that the tangent lines have slopes which match my calculations.

Activity 1.3.7.

Calculate the implicit derivative for the following locus. Sketch the locus and observe the expected slopes; compare those with the slopes given by the implicit derivatives.

\begin{equation*} xy = 1 \end{equation*}

Solution.

\begin{align*} \frac{d}{dx} xy \amp = \frac{d}{dx} 1 \\ x \frac{dy}{dx} + y \amp = 0 \\ \frac{dy}{dx} = \frac{-y}{x} \end{align*}

By solving for \(\frac{dy}{dx}\text{,}\) I get an expression in both \(x\) and \(y\) that describes the derivative. I can evalute this at a few points.

\begin{gather*} \frac{dy}{dx} (1,1) = -1 \\ \frac{dy}{dx} (-1,-1) = -1 \\ \frac{dy}{dx} \left( 3, \frac{1}{3} \right) = \frac{-1}{9} \\ \frac{dy}{dx} \left( \frac{-1}{3}, -3 \right) = -9 \end{gather*}

Figure 1.3.4. Implicit Derivatives to \(xy = 1\)

Figure 1.3.4 shows the graph of the locus. Looking at the three points where I evaluted the derivaitve, I can see that the tangent lines have slopes which match my calculations.

Activity 1.3.8.

Calculate the implicit derivative for the following locus. Sketch the locus and observe the expected slopes; compare those with the slopes given by the implicit derivatives.

\begin{equation*} x^2 + y^2 = -3 \end{equation*}

Solution.

I could go through the process here of calculating the implicit derivative. That process would produce an expression for \(\frac{dy}{dx}\text{,}\) but the expression would be useless. This locus doesn’t have any points! Therefore, it can’t have any tangent lines.

Subsection 1.3.3 Conceptual Review Questions

Why are hyperbolics like trigonometric functions?
What is hyperbolic angle and how does it define the hyperbolic functions?
Why are there sign differences between hyperbolic and trigonometric identities?

Subsection 1.3.4 Extra Practice Activities

Activity 1.3.9.

Calculate these limits.

\begin{equation*} \lim_{x \rightarrow \infty} \frac{\sinh x}{x^4} \end{equation*}
\begin{equation*} \lim_{x \rightarrow 0} \frac{\sin x}{\sinh x} \end{equation*}

Solution.

\begin{equation*} \lim_{x \rightarrow \infty} \frac{\sinh x}{x^4} = \lim_{x \rightarrow \infty} \frac{e^x - e^{-x}}{2x^4} = \infty \end{equation*}

I can change the hyperbolics into their exponential form and use asymptotic analysis. Then I see the numerator has higher asymptotic order.
\begin{equation*} \lim_{x \rightarrow 0} \frac{\sin x}{\sinh x} = \lim_{x \rightarrow 0} \frac{\sin x}{x} \lim_{x \rightarrow 0} \frac{x}{\sinh x} = 1 \cdot 1 = 1 \end{equation*}

This is a neat trick. By multiplying by \(\frac{x}{x}\text{,}\) which doesn’t change the fraction, I can split this up and use the known limits for \(\frac{\sin x}{x}\) and \(\frac{\sinh x}{x}\text{.}\)

Activity 1.3.10.

Calculate these derivatives.

\begin{equation*} \frac{d}{dx} \sinh 2x \coth x \end{equation*}
\begin{equation*} \frac{d}{dx} \frac{\sinh^2 x}{e^x} \end{equation*}

Solution.

\begin{align*} \frac{d}{dx} \sinh 2x \coth x \amp = \frac{d}{dx} 2 \sinh x \cosh x \frac{\cosh x}{\sinh x} \\ \amp = 2 \frac{d}{dx} \cosh^2 x = 4 \cosh x \frac{d}{dx} \cosh x = 4 \cosh x \sinh x \end{align*}

This is a product rule derivative.
\begin{align*} \frac{d}{dx} \frac{\sinh^2 x}{e^x} \amp = \frac{d}{dx} \frac{e^{2x} - 2 + e^{-2x}}{4e^x} = \frac{1}{4} \frac{d}{dx} e^x - 2e^{-x} + e^{-3x} \\ \amp = \frac{1}{4} \left( e^x + 2e^{-x} - 3e^{-3x} \right) \end{align*}

I change the hyperbolic into an exponential, since there was an exponential in the denominator. That changed this into three straightforward derivatives. In any exponential with more than just \(x\) in the exponent, I use the chain rule for the derivative.

Activity 1.3.11.

Calculate these integrals.

\begin{equation*} \int_{\ln 2}^{\ln 4} \cosh x dx \end{equation*}
\begin{equation*} \int_0^{\ln 5} \sinh^2 x + \cosh^2 x dx \end{equation*}

Solution.

\begin{align*} \int_{\ln 2}^{\ln 4} \cosh x dx \amp = \left. \sinh x \right|_{\ln 2}^{\ln 4} = \frac{e^{\ln 4} - e^{-\ln 4} - e^{\ln 2}+ e^{-\ln 2}}{2} \\ \amp = \frac{4 - \frac{1}{4} - 2 + \frac{1}{2}}{2} = \frac{9}{8} \end{align*}

This is a direct antiderivative. After the antiderivative, I use the exponential form to evaluate on the bounds.
\begin{align*} \int_0^{\ln 5} \sinh^2 x + \cosh^2 x dx \amp = \int_0^{\ln 5} \cosh (2x) = \left. \frac{1}{2} \sinh (2x) \right|_0^{\ln 5} \\ \amp = \frac{1}{2} \frac{e^{2\ln 5} - e^{-2\ln 5}}{2} - 0 = \frac{1}{4} \left( 25 - \frac{1}{25} \right) \\ \amp = \frac{624}{100} = \frac{156}{25} \end{align*}

I use a identity to turn this into a fairly direct integral. (To deal with the \((2x)\) inside the hyperbolic cosine, I use a substitution \(u = 2x\text{.}\)) I use the exponential form to evaluate on the bounds.

Prev Top Next