Menu Close

Question-197057

Tinku Tara 0 Comments
Pin




Question Number 197057 by ajfour last updated on 07/Sep/23
Commented by ajfour last updated on 07/Sep/23
The ant has to climb up the plane and surmount the wall of height c, and descend then reach B. Find the shortest length of path.
$${The}\:{ant}\:{has}\:{to}\:{climb}\:{up}\:{the}\:{plane} \\ $$$${and}\:{surmount}\:{the}\:{wall}\:{of}\:{height}\:{c}, \\ $$$${and}\:{descend}\:{then}\:{reach}\:{B}.\:{Find}\:{the} \\ $$$${shortest}\:{length}\:{of}\:{path}. \\ $$
Commented by Frix last updated on 08/Sep/23
Depending on the values of a, b, c the path along the diagonal of the rectangle ab and the rectangle bc on the right side can be shorter. We have (√(b^2 +(a+b+c)^2 ))⋛(√(a^2 +b^2 ))+(√(b^2 +c^2 )) Let b=pa∧c=qa; p, q >0 a(√(p^2 +(1+p+q)^2 ))⋛a(√(1+p^2 ))+a(√(p^2 +q^2 )) p^2 +(p+q+1)^2 ⋛2p^2 +q^2 +1+2(√((p^2 +q^2 )(p^2 +1))) p+pq+q⋛(√((p^2 +q^2 )(p^2 +1))) p(p^3 −2pq−2q(q+1))⋚0 p^3 −2pq−2q(q+1)⋚0 Let q=αp; α>0 p(p^2 −2α(α+1)p−2α)⋚0 p^2 −2α(α+1)p−2α⋚0 p^2 −2α(α+1)p−2α=0 p=α(α+1)±(√(α(α+2)(α^2 +1))) p_1 =α(α+1)−(√(α(α+2)(α^2 +1)))<0∀α>0 p_2 =α(α+1)+(√(α(α+2)(α^2 +1)))>0∀α>0 We have p>0 ⇒ (p−p_1 )(p−p_2 )<0; 0<p<p_2 (√(b^2 +(a+b+c)^2 ))>(√(a^2 +b^2 ))+(√(b^2 +c^2 )) (p−p_1 )(p−p_2 )=0; p=p_2 (√(b^2 +(a+b+c)^2 ))=(√(a^2 +b^2 ))+(√(b^2 +c^2 )) (p−p_1 )(p−p_2 )>0; p>p_2 (√(b^2 +(a+b+c)^2 ))<(√(a^2 +b^2 ))+(√(b^2 +c^2 ))
$$\mathrm{Depending}\:\mathrm{on}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{a},\:{b},\:{c}\:\mathrm{the}\:\mathrm{path} \\ $$$$\mathrm{along}\:\mathrm{the}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{the}\:\mathrm{rectangle}\:{ab}\:\mathrm{and} \\ $$$$\mathrm{the}\:\mathrm{rectangle}\:{bc}\:\mathrm{on}\:\mathrm{the}\:\mathrm{right}\:\mathrm{side}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{shorter}.\:\mathrm{We}\:\mathrm{have} \\ $$$$\sqrt{{b}^{\mathrm{2}} +\left({a}+{b}+{c}\right)^{\mathrm{2}} }\lesseqgtr\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$ \\ $$$$\mathrm{Let}\:{b}={pa}\wedge{c}={qa};\:{p},\:{q}\:>\mathrm{0} \\ $$$${a}\sqrt{{p}^{\mathrm{2}} +\left(\mathrm{1}+{p}+{q}\right)^{\mathrm{2}} }\lesseqgtr{a}\sqrt{\mathrm{1}+{p}^{\mathrm{2}} }+{a}\sqrt{{p}^{\mathrm{2}} +{q}^{\mathrm{2}} } \\ $$$${p}^{\mathrm{2}} +\left({p}+{q}+\mathrm{1}\right)^{\mathrm{2}} \lesseqgtr\mathrm{2}{p}^{\mathrm{2}} +{q}^{\mathrm{2}} +\mathrm{1}+\mathrm{2}\sqrt{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${p}+{pq}+{q}\lesseqgtr\sqrt{\left({p}^{\mathrm{2}} +{q}^{\mathrm{2}} \right)\left({p}^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$${p}\left({p}^{\mathrm{3}} −\mathrm{2}{pq}−\mathrm{2}{q}\left({q}+\mathrm{1}\right)\right)\gtreqless\mathrm{0} \\ $$$${p}^{\mathrm{3}} −\mathrm{2}{pq}−\mathrm{2}{q}\left({q}+\mathrm{1}\right)\gtreqless\mathrm{0} \\ $$$$ \\ $$$$\mathrm{Let}\:{q}=\alpha{p};\:\alpha>\mathrm{0} \\ $$$${p}\left({p}^{\mathrm{2}} −\mathrm{2}\alpha\left(\alpha+\mathrm{1}\right){p}−\mathrm{2}\alpha\right)\gtreqless\mathrm{0} \\ $$$${p}^{\mathrm{2}} −\mathrm{2}\alpha\left(\alpha+\mathrm{1}\right){p}−\mathrm{2}\alpha\gtreqless\mathrm{0} \\ $$$$ \\ $$$$\:\:\:\:\:{p}^{\mathrm{2}} −\mathrm{2}\alpha\left(\alpha+\mathrm{1}\right){p}−\mathrm{2}\alpha=\mathrm{0} \\ $$$$\:\:\:\:\:{p}=\alpha\left(\alpha+\mathrm{1}\right)\pm\sqrt{\alpha\left(\alpha+\mathrm{2}\right)\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)} \\ $$$$\:\:\:\:\:{p}_{\mathrm{1}} =\alpha\left(\alpha+\mathrm{1}\right)−\sqrt{\alpha\left(\alpha+\mathrm{2}\right)\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}<\mathrm{0}\forall\alpha>\mathrm{0} \\ $$$$\:\:\:\:\:{p}_{\mathrm{2}} =\alpha\left(\alpha+\mathrm{1}\right)+\sqrt{\alpha\left(\alpha+\mathrm{2}\right)\left(\alpha^{\mathrm{2}} +\mathrm{1}\right)}>\mathrm{0}\forall\alpha>\mathrm{0} \\ $$$$ \\ $$$$\mathrm{We}\:\mathrm{have}\:{p}>\mathrm{0}\:\Rightarrow \\ $$$$\left({p}−{p}_{\mathrm{1}} \right)\left({p}−{p}_{\mathrm{2}} \right)<\mathrm{0};\:\mathrm{0}<{p}<{p}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\sqrt{{b}^{\mathrm{2}} +\left({a}+{b}+{c}\right)^{\mathrm{2}} }>\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\left({p}−{p}_{\mathrm{1}} \right)\left({p}−{p}_{\mathrm{2}} \right)=\mathrm{0};\:{p}={p}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\sqrt{{b}^{\mathrm{2}} +\left({a}+{b}+{c}\right)^{\mathrm{2}} }=\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$$$\left({p}−{p}_{\mathrm{1}} \right)\left({p}−{p}_{\mathrm{2}} \right)>\mathrm{0};\:{p}>{p}_{\mathrm{2}} \\ $$$$\:\:\:\:\:\sqrt{{b}^{\mathrm{2}} +\left({a}+{b}+{c}\right)^{\mathrm{2}} }<\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }+\sqrt{{b}^{\mathrm{2}} +{c}^{\mathrm{2}} } \\ $$
Answered by Frix last updated on 07/Sep/23
(√(b^2 +(a+b+c)^2 )) The shortest path is the diagonal of the rectangle we get by unfolding the object. Its sides are b and (a+b+c)
$$\sqrt{{b}^{\mathrm{2}} +\left({a}+{b}+{c}\right)^{\mathrm{2}} } \\ $$$$\mathrm{The}\:\mathrm{shortest}\:\mathrm{path}\:\mathrm{is}\:\mathrm{the}\:\mathrm{diagonal}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{rectangle}\:\mathrm{we}\:\mathrm{get}\:\mathrm{by}\:\mathrm{unfolding}\:\mathrm{the}\:\mathrm{object}. \\ $$$$\mathrm{Its}\:\mathrm{sides}\:\mathrm{are}\:{b}\:\mathrm{and}\:\left({a}+{b}+{c}\right) \\ $$
Answered by mahdipoor last updated on 07/Sep/23
AB=AM+MN+NB⇒ p(x,y)=(√(a^2 +x^2 ))+(√(c^2 +(b−x−y)^2 ))+(√(y^2 +b^2 )) 0≤x,y≤b (dp/dx)=0=(x/( (√(x^2 +a^2 ))))+(((−b+x+y))/( (√(c^2 +(b−x−y)^2 )))) (dp/dy)=0=(y/( (√(y^2 +b^2 ))))+(((−b+x+y))/( (√(c^2 +(b−x−y)^2 )))) ⇒(x/( (√(x^2 +a^2 ))))=(y/( (√(y^2 +b^2 ))))⇒bx=ay (x/( (√(x^2 +a^2 ))))=((b−(1+(b/a))x)/( (√(c^2 +(b−(1+(b/a))x)^2 ))))⇒ x^2 (c^2 +(b−mx)^2 )=(x^2 +a^2 )(b−mx)^2 ⇒x^2 c^2 =a^2 (b−mx)^2 ⇒xc=ab−(a+b)x ⇒x=((ab)/((a+b)+c)) & y=(b^2 /((a+b)+c)) 0≤x,y≤b
$${AB}={AM}+{MN}+{NB}\Rightarrow \\ $$$${p}\left({x},{y}\right)=\sqrt{{a}^{\mathrm{2}} +{x}^{\mathrm{2}} }+\sqrt{{c}^{\mathrm{2}} +\left({b}−{x}−{y}\right)^{\mathrm{2}} }+\sqrt{{y}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$\mathrm{0}\leqslant{x},{y}\leqslant{b} \\ $$$$\frac{{dp}}{{dx}}=\mathrm{0}=\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}+\frac{\left(−{b}+{x}+{y}\right)}{\:\sqrt{{c}^{\mathrm{2}} +\left({b}−{x}−{y}\right)^{\mathrm{2}} }} \\ $$$$\frac{{dp}}{{dy}}=\mathrm{0}=\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +{b}^{\mathrm{2}} }}+\frac{\left(−{b}+{x}+{y}\right)}{\:\sqrt{{c}^{\mathrm{2}} +\left({b}−{x}−{y}\right)^{\mathrm{2}} }} \\ $$$$\Rightarrow\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}=\frac{{y}}{\:\sqrt{{y}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\Rightarrow{bx}={ay} \\ $$$$\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +{a}^{\mathrm{2}} }}=\frac{{b}−\left(\mathrm{1}+\frac{{b}}{{a}}\right){x}}{\:\sqrt{{c}^{\mathrm{2}} +\left({b}−\left(\mathrm{1}+\frac{{b}}{{a}}\right){x}\right)^{\mathrm{2}} }}\Rightarrow \\ $$$${x}^{\mathrm{2}} \left({c}^{\mathrm{2}} +\left({b}−{mx}\right)^{\mathrm{2}} \right)=\left({x}^{\mathrm{2}} +{a}^{\mathrm{2}} \right)\left({b}−{mx}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} {c}^{\mathrm{2}} ={a}^{\mathrm{2}} \left({b}−{mx}\right)^{\mathrm{2}} \Rightarrow{xc}={ab}−\left({a}+{b}\right){x} \\ $$$$\Rightarrow{x}=\frac{{ab}}{\left({a}+{b}\right)+{c}}\:\:\:\:\&\:\:\:{y}=\frac{{b}^{\mathrm{2}} }{\left({a}+{b}\right)+{c}}\:\:\:\:\:\:\mathrm{0}\leqslant{x},{y}\leqslant{b} \\ $$
Commented by Frix last updated on 07/Sep/23
You don′t need this. Look at my explanation.
$$\mathrm{You}\:\mathrm{don}'\mathrm{t}\:\mathrm{need}\:\mathrm{this}.\:\mathrm{Look}\:\mathrm{at}\:\mathrm{my}\:\mathrm{explanation}. \\ $$
Commented by mahdipoor last updated on 07/Sep/23
Thats right your solution is logical and faster, the answer is the same for both
$$\mathrm{Thats}\:\mathrm{right}\:\mathrm{your}\:\mathrm{solution}\:\mathrm{is}\:\mathrm{logical}\:\mathrm{and} \\ $$$$\mathrm{faster},\:\mathrm{the}\:\mathrm{answer}\:\mathrm{is}\:\mathrm{the}\:\mathrm{same}\:\mathrm{for}\:\mathrm{both} \\ $$
Commented by Frix last updated on 07/Sep/23
Yes.
$$\mathrm{Yes}. \\ $$
Answered by ajfour last updated on 08/Sep/23

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *