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Question-187362

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Question Number 187362 by ajfour last updated on 16/Feb/23
Commented by ajfour last updated on 16/Feb/23
blue curve: y=x^3 −x black one: y=x^3 −x+k Find the equation of the shown common tangent.
$${blue}\:{curve}:\:\:{y}={x}^{\mathrm{3}} −{x} \\ $$$${black}\:{one}:\:\:\:\:{y}={x}^{\mathrm{3}} −{x}+{k} \\ $$$${Find}\:{the}\:{equation}\:{of}\:{the}\:{shown} \\ $$$${common}\:{tangent}. \\ $$
Answered by a.lgnaoui last updated on 17/Feb/23
Blue[curve y_1 =x^3 −x the equation of tangente is (dy/dx)=3x^2 −1 y has mean[and max to x: x_(1m) =(1/( (√3))) and x_(1max) =((−1)/( (√3))) point M_(12) (((−1)/( (√3))),(2/( 3(√3)))) N_(12) ((1/( (√3))),((−2)/(3(√3)))) the red line tengente also to blue and black[curve y_2 =x^3 −x+k same[(x min,x max) (((−1)/( (√3))),(2/( (√3)))+k) ,((1/( (√3))),k−(2/( (√3)))) y_2 (0)=k the equation y=ax+b of tengente must verifie: M_1 (((−1)/( (√3))),(2/(3(√3)))) and N((1/( (√3))),k−(2/( (√3)))) { (((2/( (√3))) =((−1)/( (√3)))a+b (1))),((k−(2/( (√3)))=(a/( (√3)))+b (2))) :} (1)+(2) ⇒k=2b b=(k/2). (2)−(1) ⇒k−(4/( (√3)))=((2a)/( (√3))) a=(k/2)(√3) −2 •Equation of tengente: y=((k/2)(√3) −2)x+(k/2) with k=y_2 (1) ; k>1
$${Blue}\left[{curve}\:\:\:\:\:{y}_{\mathrm{1}} ={x}^{\mathrm{3}} −{x}\right. \\ $$$${the}\:{equation}\:{of}\:\:{tangente}\:{is} \\ $$$$\frac{{dy}}{{dx}}=\mathrm{3}{x}^{\mathrm{2}} −\mathrm{1}\:\:{y}\:{has}\:{mean}\left[{and}\:{max}\:{to}\:{x}:\right. \\ $$$${x}_{\mathrm{1}{m}} =\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\:\:\:{and}\:{x}_{\mathrm{1}{max}} =\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}} \\ $$$${point}\:{M}_{\mathrm{12}} \left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}},\frac{\mathrm{2}}{\:\mathrm{3}\sqrt{\mathrm{3}}}\right)\:\:{N}_{\mathrm{12}} \left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},\frac{−\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right) \\ $$$${the}\:{red}\:\:{line}\:{tengente}\:{also}\:{to} \\ $$$${blue}\:{and}\:{black}\left[{curve}\right. \\ $$$${y}_{\mathrm{2}} ={x}^{\mathrm{3}} −{x}+{k}\:\:\:{same}\left[\left({x}\:{min},{x}\:{max}\right)\right. \\ $$$$\:\:\left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}},\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}+{k}\right)\:\:\:\:,\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},{k}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$${y}_{\mathrm{2}} \left(\mathrm{0}\right)={k} \\ $$$${the}\:{equation}\:\boldsymbol{{y}}=\boldsymbol{{ax}}+\boldsymbol{{b}}\:{of}\:{tengente} \\ $$$${must}\:{verifie}: \\ $$$${M}_{\mathrm{1}} \left(\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}},\frac{\mathrm{2}}{\mathrm{3}\sqrt{\mathrm{3}}}\right)\:\:\:\:{and}\:\:\:{N}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}},{k}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\right) \\ $$$$\begin{cases}{\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\:\:\:\:\:\:\:=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}{a}+{b}\:\:\:\:\left(\mathrm{1}\right)}\\{{k}−\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}=\frac{{a}}{\:\sqrt{\mathrm{3}}}+{b}\:\:\:\:\:\:\:\left(\mathrm{2}\right)}\end{cases} \\ $$$$\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:\:\:\:\Rightarrow{k}=\mathrm{2}{b}\:\:\:\:\:\:\:\boldsymbol{{b}}=\frac{\boldsymbol{{k}}}{\mathrm{2}}. \\ $$$$\left(\mathrm{2}\right)−\left(\mathrm{1}\right)\:\:\:\:\Rightarrow{k}−\frac{\mathrm{4}}{\:\sqrt{\mathrm{3}}}=\frac{\mathrm{2}{a}}{\:\sqrt{\mathrm{3}}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\boldsymbol{{a}}=\frac{{k}}{\mathrm{2}}\sqrt{\mathrm{3}}\:−\mathrm{2} \\ $$$$\bullet\boldsymbol{{Equation}}\:\boldsymbol{{of}}\:\boldsymbol{{tengente}}: \\ $$$$\:\:\:\:\:\:\boldsymbol{{y}}=\left(\frac{\boldsymbol{{k}}}{\mathrm{2}}\sqrt{\mathrm{3}}\:−\mathrm{2}\right)\boldsymbol{{x}}+\frac{\boldsymbol{{k}}}{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{with}\:\:{k}={y}_{\mathrm{2}} \left(\mathrm{1}\right)\:\:;\:\:\:\:\:\:{k}>\mathrm{1} \\ $$
Commented by a.lgnaoui last updated on 17/Feb/23
k∈R not specified! tengente[here dependant of y_1 ,y_2
$${k}\in\mathbb{R}\:\:\:\:{not}\:{specified}! \\ $$$${tengente}\left[{here}\:{dependant}\right. \\ $$$${of}\:{y}_{\mathrm{1}} ,{y}_{\mathrm{2}} \:\: \\ $$
Answered by ajfour last updated on 17/Feb/23
Tangent y=mx+(k/2) (symmetry!) Intersection?with curve y=x^3 −x=mx+(k/2) x^3 −x=mx+(k/2) x^3 −(m+1)x−(k/2)=0 tangency ⇒double root ⇒ D=0 (k^2 /(16))=(((m+1)^3 )/(27)) ⇒ m=(3/4)(4k^2 )^(1/3) −1 Hence eq. common tangent y=((3/4)(4k^2 )^(1/3) −1)x+(k/2)
$${Tangent} \\ $$$${y}={mx}+\frac{{k}}{\mathrm{2}}\:\:\:\:\left({symmetry}!\right) \\ $$$${Intersection}?{with}\:{curve}\:\: \\ $$$${y}={x}^{\mathrm{3}} −{x}={mx}+\frac{{k}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −{x}={mx}+\frac{{k}}{\mathrm{2}} \\ $$$${x}^{\mathrm{3}} −\left({m}+\mathrm{1}\right){x}−\frac{{k}}{\mathrm{2}}=\mathrm{0} \\ $$$${tangency}\:\Rightarrow{double}\:{root}\:\Rightarrow\:{D}=\mathrm{0} \\ $$$$\frac{{k}^{\mathrm{2}} }{\mathrm{16}}=\frac{\left({m}+\mathrm{1}\right)^{\mathrm{3}} }{\mathrm{27}} \\ $$$$\Rightarrow\:\:{m}=\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{4}{k}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} −\mathrm{1} \\ $$$${Hence}\:{eq}.\:{common}\:\:{tangent} \\ $$$${y}=\left(\frac{\mathrm{3}}{\mathrm{4}}\left(\mathrm{4}{k}^{\mathrm{2}} \right)^{\mathrm{1}/\mathrm{3}} −\mathrm{1}\right){x}+\frac{{k}}{\mathrm{2}} \\ $$

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