Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 102804 by bramlex last updated on 11/Jul/20

The coordinates of two points A  & B are (0,8) and (9,4)  respectively. The point P with  coordinate (p,0) lies on the  x−axis where 0<p<9. Let s  denotes the sum of the length of  two segments PA and PB . by  expressing s in terms of p  otherwise, show that (ds/dp) =  (p/(√(64+p^2 ))) − ((9−p)/(√(16+(9−p)^2 )))

$$\mathrm{The}\:\mathrm{coordinates}\:\mathrm{of}\:\mathrm{two}\:\mathrm{points}\:\mathrm{A} \\ $$ $$\&\:\mathrm{B}\:\mathrm{are}\:\left(\mathrm{0},\mathrm{8}\right)\:\mathrm{and}\:\left(\mathrm{9},\mathrm{4}\right) \\ $$ $$\mathrm{respectively}.\:\mathrm{The}\:\mathrm{point}\:\mathrm{P}\:\mathrm{with} \\ $$ $$\mathrm{coordinate}\:\left(\mathrm{p},\mathrm{0}\right)\:\mathrm{lies}\:\mathrm{on}\:\mathrm{the} \\ $$ $$\mathrm{x}−\mathrm{axis}\:\mathrm{where}\:\mathrm{0}<\mathrm{p}<\mathrm{9}.\:\mathrm{Let}\:\mathrm{s} \\ $$ $$\mathrm{denotes}\:\mathrm{the}\:\mathrm{sum}\:\mathrm{of}\:\mathrm{the}\:\mathrm{length}\:\mathrm{of} \\ $$ $$\mathrm{two}\:\mathrm{segments}\:\mathrm{PA}\:\mathrm{and}\:\mathrm{PB}\:.\:\mathrm{by} \\ $$ $$\mathrm{expressing}\:\mathrm{s}\:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{p} \\ $$ $$\mathrm{otherwise},\:\mathrm{show}\:\mathrm{that}\:\frac{\mathrm{ds}}{\mathrm{dp}}\:= \\ $$ $$\frac{\mathrm{p}}{\sqrt{\mathrm{64}+\mathrm{p}^{\mathrm{2}} }}\:−\:\frac{\mathrm{9}−\mathrm{p}}{\sqrt{\mathrm{16}+\left(\mathrm{9}−\mathrm{p}\right)^{\mathrm{2}} }} \\ $$

Answered by bemath last updated on 11/Jul/20

s = PA +PB = (√(p^2 +64))  +(√((9−p)^2 +16))  (ds/dp) = ((2p)/(2(√(64+p^2 )))) +  ((2(9−p)(−1))/(2(√(16+(9−p)^2 )))) =  (p/(√(64+p^2 )))−(((9−p))/(√(16+(9−p)^2 ))) □

$${s}\:=\:{PA}\:+{PB}\:=\:\sqrt{{p}^{\mathrm{2}} +\mathrm{64}} \\ $$ $$+\sqrt{\left(\mathrm{9}−{p}\right)^{\mathrm{2}} +\mathrm{16}} \\ $$ $$\frac{{ds}}{{dp}}\:=\:\frac{\mathrm{2}{p}}{\mathrm{2}\sqrt{\mathrm{64}+{p}^{\mathrm{2}} }}\:+ \\ $$ $$\frac{\mathrm{2}\left(\mathrm{9}−{p}\right)\left(−\mathrm{1}\right)}{\mathrm{2}\sqrt{\mathrm{16}+\left(\mathrm{9}−{p}\right)^{\mathrm{2}} }}\:= \\ $$ $$\frac{{p}}{\sqrt{\mathrm{64}+{p}^{\mathrm{2}} }}−\frac{\left(\mathrm{9}−{p}\right)}{\sqrt{\mathrm{16}+\left(\mathrm{9}−{p}\right)^{\mathrm{2}} }}\:\square \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com