A golfer hits a golf ball at an angle of \( \theta=24^{\circ} \) with initial velocity \( v=110 \mathrm{ft} / \mathrm{s} \). (a) Estimate \( \Delta s \) if the ball is hit the same velocity but the angle increased by \( 3^{\circ} \). (b) Estimate \( \Delta s \) if the ball is hit the same angle but the velocity increased by 4 . (Give your answer to four decimal places.) (a) \( \Delta s=\square \) (b) \( \Delta s=\square \)

We start with the formula for the range of a projectile (ignoring air resistance) \[ s=\frac{v^2\sin(2\theta)}{g}, \] where • \( v \) is the initial speed, • \( \theta \) is the launch angle, and • \( g \) is the acceleration due to gravity (we use \( g=32.2\; \mathrm{ft/s^2} \)). We use the method of linear approximation. A small change in \( s \) from a change in \( \theta \) or \( v \) is given (keeping the other variable fixed) by \[ \Delta s\approx\frac{\partial s}{\partial \theta}\Delta\theta \quad \text{or} \quad \Delta s\approx\frac{\partial s}{\partial v}\Delta v, \] respectively. --- ### (a) Change in Angle by \( 3^\circ \) The partial derivative of \( s \) with respect to \( \theta \) is \[ \frac{\partial s}{\partial \theta}=\frac{v^2\cdot 2\cos(2\theta)}{g}. \] At our given values, \( v=110\; \mathrm{ft/s} \) and \( \theta=24^\circ \), we have \( 2\theta=48^\circ \). Thus \[ \frac{\partial s}{\partial \theta}=\frac{110^2\cdot 2\cos(48^\circ)}{32.2}. \] Calculating step by step: 1. \( 110^2=12100 \). 2. \( \cos(48^\circ)\approx0.6691 \). 3. Then, \[ \frac{\partial s}{\partial \theta}\approx\frac{12100\cdot2\cdot0.6691}{32.2}=\frac{12100\cdot1.3382}{32.2}. \] 4. Multiply the numerator: \[ 12100\cdot1.3382\approx16195. \] 5. Divide by \( 32.2 \): \[ \frac{16195}{32.2}\approx503.1. \] Since the change in angle is \( 3^\circ \) we must convert it to radians: \[ \Delta\theta=3^\circ\times\frac{\pi}{180}=\frac{\pi}{60}\approx0.05236\, \text{radians}. \] Thus the change in distance is approximately \[ \Delta s\approx503.1\times\frac{\pi}{60}\approx503.1\times0.05236\approx26.3280\; \mathrm{ft}. \] --- ### (b) Change in Velocity by \( 4 \; \mathrm{ft/s} \) Next, we take the partial derivative of \( s \) with respect to \( v \): \[ \frac{\partial s}{\partial v}=\frac{2v\sin(2\theta)}{g}. \] At \( v=110\; \mathrm{ft/s} \) and \( 2\theta=48^\circ \) (with \(\sin(48^\circ)\approx0.7431\)), we have \[ \frac{\partial s}{\partial v}=\frac{2\cdot110\cdot0.7431}{32.2}. \] Step by step: 1. \( 2\cdot 110=220 \). 2. \( 220\cdot0.7431\approx163.482 \). 3. Then, \[ \frac{\partial s}{\partial v}\approx\frac{163.482}{32.2}\approx5.077. \] For an increase of \( \Delta v=4\; \mathrm{ft/s} \): \[ \Delta s\approx5.077\times4\approx20.3080\; \mathrm{ft}. \] (To four decimal places, \(20.3080\) ft.) --- ### Final Answers (a) \(\Delta s\approx26.3280\; \mathrm{ft}\) (b) \(\Delta s\approx20.3080\; \mathrm{ft}\)