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Question Number 208052 by necx122 last updated on 03/Jun/24
Sketch the curve y = x^3 . (a) Find the equation of the tangent to the curve at A(1,1). (b) Find the coordinates of point B, where the tangent meets the curve again. (c) Calculate the area between the tangent B and the arc AB of the curve.
$${Sketch}\:{the}\:{curve}\:{y}\:=\:{x}^{\mathrm{3}} . \\ $$$$\left({a}\right)\:{Find}\:{the}\:{equation}\:{of}\:{the}\:{tangent} \\ $$$${to}\:{the}\:{curve}\:{at}\:{A}\left(\mathrm{1},\mathrm{1}\right). \\ $$$$\left({b}\right)\:{Find}\:{the}\:{coordinates}\:{of}\:{point}\:{B}, \\ $$$${where}\:{the}\:{tangent}\:{meets}\:{the}\:{curve}\:{again}. \\ $$$$\left({c}\right)\:{Calculate}\:{the}\:{area}\:{between}\:{the} \\ $$$${tangent}\:{B}\:{and}\:{the}\:{arc}\:{AB}\:{of}\:{the}\:{curve}. \\ $$
Commented by Frix last updated on 03/Jun/24
Tangent t: y=ax+b f: y=x^3 ⇒ f′: y=3x^2 P∈f: P= ((p),(p^3 ) ) P∈t: p^3 =ap+b but a=3p^2 ⇒ b=−2p^3 ⇒ t: y=3p^2 x−2p^3 f(x)=x^3 t(x)=3p^2 x−2p^3 d(x)=f(x)−t(x)=x^3 −3p^2 x+2p^3 d(x)=(x−p)^2 (x+2p) This has a double zero at x=p and a single zero at x=−2p. Its extreme is at x=−p with d(−p)=4p^3 ⇒ for p<0 we have a local min, for p>0 a local max ⇒ 4p^3 ≤d(x)≤0; p<0∧p≤x≤−2p ⇒ Area is −∫_p ^(−2p) d(x)dx=∫_(−2p) ^p d(x)dx 0≤d(x)≤4p^3 ; p>0∧−2p≤x≤p ⇒ Area is ∫_(−2p) ^p d(x)dx ⇒ Area for p≠0 is always ∫_(−2p) ^p d(x)dx
$$\mathrm{Tangent}\:{t}:\:{y}={ax}+{b} \\ $$$${f}:\:{y}={x}^{\mathrm{3}} \:\Rightarrow\:{f}':\:{y}=\mathrm{3}{x}^{\mathrm{2}} \\ $$$${P}\in{f}:\:{P}=\begin{pmatrix}{{p}}\\{{p}^{\mathrm{3}} }\end{pmatrix} \\ $$$${P}\in{t}:\:{p}^{\mathrm{3}} ={ap}+{b}\:\mathrm{but}\:{a}=\mathrm{3}{p}^{\mathrm{2}} \:\Rightarrow\:{b}=−\mathrm{2}{p}^{\mathrm{3}} \:\Rightarrow \\ $$$${t}:\:{y}=\mathrm{3}{p}^{\mathrm{2}} {x}−\mathrm{2}{p}^{\mathrm{3}} \\ $$$$ \\ $$$${f}\left({x}\right)={x}^{\mathrm{3}} \\ $$$${t}\left({x}\right)=\mathrm{3}{p}^{\mathrm{2}} {x}−\mathrm{2}{p}^{\mathrm{3}} \\ $$$${d}\left({x}\right)={f}\left({x}\right)−{t}\left({x}\right)={x}^{\mathrm{3}} −\mathrm{3}{p}^{\mathrm{2}} {x}+\mathrm{2}{p}^{\mathrm{3}} \\ $$$${d}\left({x}\right)=\left({x}−{p}\right)^{\mathrm{2}} \left({x}+\mathrm{2}{p}\right) \\ $$$$\mathrm{This}\:\mathrm{has}\:\mathrm{a}\:\mathrm{double}\:\mathrm{zero}\:\mathrm{at}\:{x}={p}\:\mathrm{and}\:\mathrm{a}\:\mathrm{single} \\ $$$$\mathrm{zero}\:\mathrm{at}\:{x}=−\mathrm{2}{p}.\:\mathrm{Its}\:\mathrm{extreme}\:\mathrm{is}\:\mathrm{at}\:{x}=−{p} \\ $$$$\mathrm{with}\:{d}\left(−{p}\right)=\mathrm{4}{p}^{\mathrm{3}} \:\Rightarrow\:\mathrm{for}\:{p}<\mathrm{0}\:\mathrm{we}\:\mathrm{have}\:\mathrm{a} \\ $$$$\mathrm{local}\:\mathrm{min},\:\mathrm{for}\:{p}>\mathrm{0}\:\mathrm{a}\:\mathrm{local}\:\mathrm{max}\:\Rightarrow \\ $$$$\mathrm{4}{p}^{\mathrm{3}} \leqslant{d}\left({x}\right)\leqslant\mathrm{0};\:{p}<\mathrm{0}\wedge{p}\leqslant{x}\leqslant−\mathrm{2}{p} \\ $$$$\Rightarrow\:\mathrm{Area}\:\mathrm{is}\:\:−\underset{{p}} {\overset{−\mathrm{2}{p}} {\int}}{d}\left({x}\right){dx}=\underset{−\mathrm{2}{p}} {\overset{{p}} {\int}}{d}\left({x}\right){dx} \\ $$$$\mathrm{0}\leqslant{d}\left({x}\right)\leqslant\mathrm{4}{p}^{\mathrm{3}} ;\:{p}>\mathrm{0}\wedge−\mathrm{2}{p}\leqslant{x}\leqslant{p} \\ $$$$\Rightarrow\:{A}\mathrm{rea}\:\mathrm{is}\:\underset{−\mathrm{2}{p}} {\overset{{p}} {\int}}{d}\left({x}\right){dx} \\ $$$$\Rightarrow\:\mathrm{Area}\:\mathrm{for}\:{p}\neq\mathrm{0}\:\mathrm{is}\:\mathrm{always}\:\underset{−\mathrm{2}{p}} {\overset{{p}} {\int}}{d}\left({x}\right){dx} \\ $$
Commented by Frix last updated on 03/Jun/24
y=x^3 y′=3x^2 Tangent at x=p: y=3p^2 x−2p^3 Intersection x^3 −(3p^2 x−2p^3 )=0 (x−p)^2 (x+2p)=0 x_1 =p (obvious) x_2 =−2p The area is ∫_(−2p) ^p (x^3 −(3p^2 x−2p^3 ))dx=((27p^4 )/4)
$${y}={x}^{\mathrm{3}} \\ $$$${y}'=\mathrm{3}{x}^{\mathrm{2}} \\ $$$$\mathrm{Tangent}\:\mathrm{at}\:{x}={p}:\:{y}=\mathrm{3}{p}^{\mathrm{2}} {x}−\mathrm{2}{p}^{\mathrm{3}} \\ $$$$\mathrm{Intersection} \\ $$$${x}^{\mathrm{3}} −\left(\mathrm{3}{p}^{\mathrm{2}} {x}−\mathrm{2}{p}^{\mathrm{3}} \right)=\mathrm{0} \\ $$$$\left({x}−{p}\right)^{\mathrm{2}} \left({x}+\mathrm{2}{p}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} ={p}\:\left(\mathrm{obvious}\right) \\ $$$${x}_{\mathrm{2}} =−\mathrm{2}{p} \\ $$$$\mathrm{The}\:\mathrm{area}\:\mathrm{is} \\ $$$$\underset{−\mathrm{2}{p}} {\overset{{p}} {\int}}\left({x}^{\mathrm{3}} −\left(\mathrm{3}{p}^{\mathrm{2}} {x}−\mathrm{2}{p}^{\mathrm{3}} \right)\right){dx}=\frac{\mathrm{27}{p}^{\mathrm{4}} }{\mathrm{4}} \\ $$
Commented by necx122 last updated on 03/Jun/24
Thank you sir frix. However, I′m bothered about how you came about the y-intercept for the tangent as −2p^3 . Also, in sketching the curve and calculating the area bounded by the arent we supposed to calculate for all possible regions both above and below the x-axis which may indicate a different value for the area bounded by the curve and the line. Thank you for your help. I always appreciate.
$${Thank}\:{you}\:{sir}\:{frix}.\:{However},\:{I}'{m} \\ $$$${bothered}\:{about}\:{how}\:{you}\:{came}\:{about} \\ $$$${the}\:{y}-{intercept}\:{for}\:{the}\:{tangent}\:{as}\:−\mathrm{2}{p}^{\mathrm{3}} . \\ $$$${Also},\:{in}\:{sketching}\:{the}\:{curve}\:{and}\: \\ $$$${calculating}\:{the}\:{area}\:{bounded}\:{by}\:{the} \\ $$$${arent}\:{we}\:{supposed}\:{to}\:{calculate}\:{for}\:{all} \\ $$$${possible}\:{regions}\:{both}\:{above}\:{and}\:{below} \\ $$$${the}\:{x}-{axis}\:{which}\:{may}\:{indicate}\:{a}\:{different} \\ $$$${value}\:{for}\:{the}\:{area}\:{bounded}\:{by}\:{the}\:{curve} \\ $$$${and}\:{the}\:{line}. \\ $$$${Thank}\:{you}\:{for}\:{your}\:{help}.\:{I}\:{always}\: \\ $$$${appreciate}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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