A ball thrown from ground level with initial speed v and at an angle $\alpha$ with the horizontal hits the ground at a distance $s = \frac { v ^ { 2 } \sin ( 2 \alpha ) } { g }$ away from the starting point (where g is the acceleration due to gravity).Calculate the differential ds.
A) $d s = \frac { 2 v \sin ( 2 \alpha ) } { g } d v + \frac { 2 v ^ { 2 } \cos ( 2 \alpha ) } { g } d \alpha$
B) $d s = 2 v \sin ( 2 \alpha ) d v + 2 v ^ { 2 } \cos ( 2 \alpha ) d \alpha$
C) $d s = \frac { 2 v \sin ( 2 \alpha ) } { g } d v + \frac { 2 v ^ { 2 } \sin ( 2 \alpha ) \cos ( 2 \alpha ) } { g } d \alpha$
D) $d s = \frac { 2 v \sin ( 2 \alpha ) } { g } d v - \frac { 2 v ^ { 2 } \cos ( 2 \alpha ) } { g } d \alpha$

Question

Quizplus · Accepted Answer

The differential $ds$ of the given function $s = \frac { v ^ { 2 } \sin ( 2 \alpha ) } { g }$ with respect to $v$ and $\alpha$ can be found using partial differentiation. The first term $\frac { 2 v \sin ( 2 \alpha ) } { g } d v$ comes from differentiating with respect to $v$, and the second term $\frac { 2 v ^ { 2 } \cos ( 2 \alpha ) } { g } d \alpha$ comes from differentiating with respect to $\alpha$, applying the chain rule for the derivative of $\sin(2\alpha)$ with respect to $\alpha$.

A Ball Thrown from Ground Level with Initial Speed V α\alphaα

A Ball Thrown from Ground Level with Initial Speed V $\alpha$