91Ó°ÊÓ

The intensity \(L(x)\) of light \(x\) meters beneath the surface of the ocean satisfies the differential equation $$\frac{d L}{d x}=-k L$$ As a diver, you know from experience that diving to 6 meters in the Caribbean Sea cuts the intensity in half. You cannot work without artificial light when the intensity falls below one-tenth of the surface value. About how deep can you expect to work without artificial light?

Use l'Hôpital's rule to find the limits. $$\lim _{\theta \rightarrow 0} \frac{3^{\sin \theta}-1}{\theta}$$

Find the derivative of \(y\) with respect to the appropriate variable. $$y=(1-\theta) \tanh ^{-1} \theta$$

Gives a formula for a function \(y=f(x) .\) In each case, find \(f^{-1}(x)\) and identify the domain and range of \(f^{-1} .\) As a check, show that \(f\left(f^{-1}(x)\right)=f^{-1}(f(x))=x\). $$f(x)=(1 / 2) x-7 / 2$$

Use l'Hôpital's rule to find the limits. $$\lim _{\theta \rightarrow 0} \frac{(1 / 2)^{\theta}-1}{\theta}$$

Find the derivative of \(y\) with respect to \(x, t,\) or \(\theta,\) as appropriate. $$y=\frac{1}{2} \ln \frac{1+x}{1-x}$$

Find \(d y / d x.\) $$\tan y=e^{x}+\ln x$$

Suppose that electricity is draining from a capacitor at a rate that is proportional to the voltage \(V\) across its terminals and that, if \(t\) is measured in seconds, $$\frac{d V}{d t}=-\frac{1}{40} V$$ Solve this equation for \(V\), using \(V_{0}\) to denote the value of \(V\) when \(t=0 .\) How long will it take the voltage to drop to \(10 \%\) of its original value?

Find the derivative of \(y\) with respect to the appropriate variable. $$y=\left(\theta^{2}+2 \theta\right) \tanh ^{-1}(\theta+1)$$

Find the derivative of \(y\) with respect to the appropriate variable. $$y=(1-t) \operatorname{coth}^{-1} \sqrt{t}$$