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Question Number 135004 by bramlexs22 last updated on 09/Mar/21
the closest distance from the point on the curve y = x ^ 3-1 to the curve x = y ^2+ 4 is equal to
$$ \\ $$the closest distance from the point on the curve y = x ^ 3-1 to the curve x = y ^2+ 4 is equal to
Commented by bramlexs22 last updated on 09/Mar/21
Commented by EDWIN88 last updated on 09/Mar/21
step(1) slope of tangent line curve y=x^3 −1 at point P(a,a^3 −1) is m_1 =3a^2 then slope of normal line is −(1/(3a^2 )) the equation of normal line at point (a,a^3 −1) is y=−(1/(3a^2 ))(x−a)+a^3 −1⇒y =−(x/(3a^2 ))+(1/(3a))+a^3 −1 step(2) slope of tangent line curve y^2 =x−4 at point Q(b^2 +4, b) is m_2 = (1/(2b)) , then slope of normal line is −2b , and the equation of normal line equal to y = −2b(x−b^2 −4)+b⇒y=−2bx+2b^3 +9b step(3) the both of equation of normal line is same , we get { ((−(1/(3a^2 )) = −2b ; a^2 b = (1/6))),(((1/(3a))+a^3 −1 = 2b^3 +9b)) :} solving for a and b , ⇒(1/(3a))+a^3 −1=2((1/(6a^2 )))^3 +9((1/(6a^2 ))) ⇒(1/(3a))+a^3 −1 = (1/(108a^6 )) +(3/(2a^2 )) 36a^5 +108a^9 −108a^6 =1+162a^4 108a^9 −108a^6 +36a^5 −162a^4 −1=0 we get { ((a=1.2067)),((b = 0.1145)) :} we get a point P(1.2067; 0.7571) and Q(4.0131; 0.1145) therefore the closest distance is ∣PQ∣ = (√((4.0131−1.2067)^2 +(0.1145−0.7571)^2 )) ∣PQ∣ ≈ 2.87903
$$\mathrm{step}\left(\mathrm{1}\right)\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{curve}\:\mathrm{y}=\mathrm{x}^{\mathrm{3}} −\mathrm{1}\:\mathrm{at}\:\mathrm{point}\:\mathrm{P}\left(\mathrm{a},\mathrm{a}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\mathrm{is}\:\mathrm{m}_{\mathrm{1}} =\mathrm{3a}^{\mathrm{2}} \:\mathrm{then}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{is}\:−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} } \\ $$$$\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line}\:\mathrm{at}\:\mathrm{point}\:\left(\mathrm{a},\mathrm{a}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$$\mathrm{is}\:\mathrm{y}=−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} }\left(\mathrm{x}−\mathrm{a}\right)+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\Rightarrow\mathrm{y}\:=−\frac{\mathrm{x}}{\mathrm{3a}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1} \\ $$$$\mathrm{step}\left(\mathrm{2}\right)\:\mathrm{slope}\:\mathrm{of}\:\mathrm{tangent}\:\mathrm{line}\:\mathrm{curve}\:\mathrm{y}^{\mathrm{2}} =\mathrm{x}−\mathrm{4}\:\mathrm{at}\:\mathrm{point} \\ $$$$\mathrm{Q}\left(\mathrm{b}^{\mathrm{2}} +\mathrm{4},\:\mathrm{b}\right)\:\mathrm{is}\:\mathrm{m}_{\mathrm{2}} \:=\:\frac{\mathrm{1}}{\mathrm{2b}}\:,\:\mathrm{then}\:\mathrm{slope}\:\mathrm{of}\:\mathrm{normal} \\ $$$$\mathrm{line}\:\mathrm{is}\:−\mathrm{2b}\:,\:\mathrm{and}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line} \\ $$$$\mathrm{equal}\:\mathrm{to}\:\mathrm{y}\:=\:−\mathrm{2b}\left(\mathrm{x}−\mathrm{b}^{\mathrm{2}} −\mathrm{4}\right)+\mathrm{b}\Rightarrow\mathrm{y}=−\mathrm{2bx}+\mathrm{2b}^{\mathrm{3}} +\mathrm{9b} \\ $$$$\mathrm{step}\left(\mathrm{3}\right)\:\mathrm{the}\:\mathrm{both}\:\mathrm{of}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{normal}\:\mathrm{line} \\ $$$$\mathrm{is}\:\mathrm{same}\:,\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} }\:=\:−\mathrm{2b}\:;\:\mathrm{a}^{\mathrm{2}} \mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{6}}}\\{\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\mathrm{2b}^{\mathrm{3}} +\mathrm{9b}}\end{cases} \\ $$$$\mathrm{solving}\:\mathrm{for}\:\mathrm{a}\:\mathrm{and}\:\mathrm{b}\:,\:\Rightarrow\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{6a}^{\mathrm{2}} }\right)^{\mathrm{3}} +\mathrm{9}\left(\frac{\mathrm{1}}{\mathrm{6a}^{\mathrm{2}} }\right) \\ $$$$\Rightarrow\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{108a}^{\mathrm{6}} }\:+\frac{\mathrm{3}}{\mathrm{2a}^{\mathrm{2}} } \\ $$$$\mathrm{36a}^{\mathrm{5}} +\mathrm{108a}^{\mathrm{9}} −\mathrm{108a}^{\mathrm{6}} =\mathrm{1}+\mathrm{162a}^{\mathrm{4}} \\ $$$$\mathrm{108a}^{\mathrm{9}} −\mathrm{108a}^{\mathrm{6}} +\mathrm{36a}^{\mathrm{5}} −\mathrm{162a}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{get}\:\begin{cases}{\mathrm{a}=\mathrm{1}.\mathrm{2067}}\\{\mathrm{b}\:=\:\mathrm{0}.\mathrm{1145}}\end{cases} \\ $$$$\mathrm{we}\:\mathrm{get}\:\mathrm{a}\:\mathrm{point}\:\mathrm{P}\left(\mathrm{1}.\mathrm{2067};\:\mathrm{0}.\mathrm{7571}\right)\:\mathrm{and}\: \\ $$$$\mathrm{Q}\left(\mathrm{4}.\mathrm{0131};\:\mathrm{0}.\mathrm{1145}\right) \\ $$$$\mathrm{therefore}\:\mathrm{the}\:\mathrm{closest}\:\mathrm{distance}\:\mathrm{is}\: \\ $$$$\mid\mathrm{PQ}\mid\:=\:\sqrt{\left(\mathrm{4}.\mathrm{0131}−\mathrm{1}.\mathrm{2067}\right)^{\mathrm{2}} +\left(\mathrm{0}.\mathrm{1145}−\mathrm{0}.\mathrm{7571}\right)^{\mathrm{2}} } \\ $$$$\mid\mathrm{PQ}\mid\:\approx\:\mathrm{2}.\mathrm{87903} \\ $$
Commented by bramlexs22 last updated on 09/Mar/21
y= x^3 −1 to x = y^2 + 4
$$\mathrm{y}=\:\mathrm{x}^{\mathrm{3}} −\mathrm{1}\:\mathrm{to}\:\mathrm{x}\:=\:\mathrm{y}^{\mathrm{2}} \:+\:\mathrm{4} \\ $$
Commented by bramlexs22 last updated on 10/Mar/21
Commented by mr W last updated on 09/Mar/21
Edwin sir: please check, i think error is here: is same , we get { ((−(1/(3a^2 )) = −2b ; a^2 b = (1/6))),(((1/(3a))+a^3 −1 = 2b^3 +9b)) :}
$${Edwin}\:{sir}: \\ $$$${please}\:{check},\:{i}\:{think}\:{error}\:{is}\:{here}: \\ $$$$\mathrm{is}\:\mathrm{same}\:,\:\mathrm{we}\:\mathrm{get}\:\begin{cases}{−\frac{\mathrm{1}}{\mathrm{3a}^{\mathrm{2}} }\:=\:−\mathrm{2b}\:;\:\mathrm{a}^{\mathrm{2}} \mathrm{b}\:=\:\frac{\mathrm{1}}{\mathrm{6}}}\\{\frac{\mathrm{1}}{\mathrm{3a}}+\mathrm{a}^{\mathrm{3}} −\mathrm{1}\:=\:\mathrm{2b}^{\mathrm{3}} +\mathrm{9b}}\end{cases} \\ $$
Commented by EDWIN88 last updated on 09/Mar/21
yes sir. thanks you sir
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by mr W last updated on 09/Mar/21
say point on curve y=x^3 −1 is P(p,p^3 −1) point on curve x=y^2 +4 is Q(q^2 +4,q) d=PQ D=d^2 =(q^2 +4−p)^2 +(q−p^3 +1)^2 (∂D/∂p)=−2(q^2 +4−p)−6p^2 (q−p^3 +1)=0 ⇒(q^2 +4−p)+3p^2 (q−p^3 +1)=0 ...(i) (∂D/∂q)=4q(q^2 +4−p)+2(q−p^3 +1)=0 ⇒2q(q^2 +4−p)+(q−p^3 +1)=0 ...(ii) ⇒2q=(1/(3p^2 )) ⇒q=(1/(6p^2 )) put this into (i): (1/(36p^4 ))+4−p+3p^2 ((1/(6p^2 ))−p^3 +1)=0 ⇒108p^9 −108p^6 +36p^5 −162p^4 −1=0 ⇒p≈1.2067 ⇒d_(min) ≈2.8791
$${say}\:{point}\:{on}\:{curve}\:{y}={x}^{\mathrm{3}} −\mathrm{1}\:{is} \\ $$$${P}\left({p},{p}^{\mathrm{3}} −\mathrm{1}\right) \\ $$$${point}\:{on}\:{curve}\:{x}={y}^{\mathrm{2}} +\mathrm{4}\:{is} \\ $$$${Q}\left({q}^{\mathrm{2}} +\mathrm{4},{q}\right) \\ $$$${d}={PQ} \\ $$$${D}={d}^{\mathrm{2}} =\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)^{\mathrm{2}} +\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\frac{\partial{D}}{\partial{p}}=−\mathrm{2}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)−\mathrm{6}{p}^{\mathrm{2}} \left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\mathrm{3}{p}^{\mathrm{2}} \left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0}\:\:\:…\left({i}\right) \\ $$$$\frac{\partial{D}}{\partial{q}}=\mathrm{4}{q}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\mathrm{2}\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{2}{q}\left({q}^{\mathrm{2}} +\mathrm{4}−{p}\right)+\left({q}−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0}\:\:\:…\left({ii}\right) \\ $$$$\Rightarrow\mathrm{2}{q}=\frac{\mathrm{1}}{\mathrm{3}{p}^{\mathrm{2}} }\:\Rightarrow{q}=\frac{\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} } \\ $$$${put}\:{this}\:{into}\:\left({i}\right): \\ $$$$\frac{\mathrm{1}}{\mathrm{36}{p}^{\mathrm{4}} }+\mathrm{4}−{p}+\mathrm{3}{p}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{6}{p}^{\mathrm{2}} }−{p}^{\mathrm{3}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow\mathrm{108}{p}^{\mathrm{9}} −\mathrm{108}{p}^{\mathrm{6}} +\mathrm{36}{p}^{\mathrm{5}} −\mathrm{162}{p}^{\mathrm{4}} −\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow{p}\approx\mathrm{1}.\mathrm{2067} \\ $$$$\Rightarrow{d}_{{min}} \approx\mathrm{2}.\mathrm{8791} \\ $$
Commented by mr W last updated on 09/Mar/21
Commented by bramlexs22 last updated on 09/Mar/21
thanks you sir
$$\mathrm{thanks}\:\mathrm{you}\:\mathrm{sir} \\ $$

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