Let $y = \sqrt{x}$. Use $dy$ for the tangent line at $x=1$ to approximate $\sqrt{2}$.

$f(x) = \sqrt{x}$, so $f'(x) = \displaystyle{\frac{1}{2\sqrt{x}}}$

$$dx = 2 - 1$$ $$dy = f'(1) \cdot 1 = \displaystyle{\frac{1}{2\sqrt{1}} \cdot 1 = \frac{1}{2}}$$So,

$$\sqrt{2} = f(1+dx) \approx f(1) + dy = \sqrt{1} + 0.5 = 1.5$$which is not too far from the actual value of $\sqrt{2} \doteq 1.414213562$.

Approximate $\sin 1$, given that $\displaystyle{\frac{\pi}{3} \approx 1.04719755}$.

$f(x) = \sin x$, so $f'(x) = \cos x$

$$dx = 1 - \displaystyle{\frac{\pi}{3}} \doteq -0.04719755$$ $$dy \doteq \cos \displaystyle{\frac{\pi}{3}} \cdot (-0.04719755) = 0.5 \cdot (-0.04719755) = -.023598775$$So,

$$\sin 1 = f\left(\frac{\pi}{3}+dx\right) \approx f \left(\displaystyle{\frac{\pi}{3}}\right) + dy = \frac{\sqrt{3}}{2} - .023598775 = 0.8424266288$$which is not too far from the actual value of $\sin 1 \doteq 0.8414709848$.

A ladder of length $L$ feet leans against a vertical wall (assume that the floor is horizontal) with the base of the ladder is $L/2$ feet from the wall. If the base of the ladder slides away from the wall by $1$ in, approximate the distance the top of the ladder slides down the wall.

$\displaystyle{\frac{\sqrt{3}}{3}}$ in

Yogi Bear is once again using balloons to steal picnic baskets from unsuspecting park visitors. This time, Yogi’s three-foot-tall friend Boo-Boo is watching this hair-brained scheme unfold from a point $30$ feet away. As the picnic basket slowly rises vertically into the air, approximate the angle Boo-Boo’s head tilts as the basket rises from $33$ feet to $34$ feet above the ground.

$\displaystyle{\frac{1}{60}}$ radians

Helium is being pumped into a spherical balloon that currently has a radius of 10 cm. Approximate by how much the surface area will increase when $0.01$ cm

^{3}more helium is pumped into the balloon.$0.002$ cm

^{2}Let $R$ be a rectangle whose sides are changing size. Suppose its length increases from $20$ inches to $21$ inches and the width decreases from $30$ inches to $29$ inches. Find both the rectangle’s approximate change in area and its actual change in area. Under what circumstances would the difference between these two values be smaller? Larger?

Approximate change: $10$ in

^{2}; actual change: $9$ in^{2}; smaller/larger changes in length and width will lead to the difference between the approximate and actual change in area being smaller/larger.Professor Hubert J. Farnsworth points a laser pointer level at a screen such that the laser beam contacts a screen $6$ meters away at an angle of $\frac{\pi}{6}$ with the plane of the screen. A tremor in Prof. Farnsworth hand causes the angle of the pointer to wobble up to $3$ degrees ($\Delta \theta = \pm \frac{\pi}{60} \approx 0.05236$ radians). Determine approximately how much the laser’s dot on the screen wobbles up and down and back and forth.

If $\theta$ is the angle the beam makes with the plane of the screen and $h$ is the height above the pointer where the laser's dot appears, then $\tan \theta = 6/h$, suggesting $h(\theta) = 6/\tan \theta$.

Now use differentials to approximate both $h(\theta + \frac{\pi}{60})$ and $h(\theta - \frac{\pi}{60})$, and subtract.