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Define-a-curve-E-by-the-parametric-equations-x-t-g-1-t-h-1-t-f-1-u-du-y-t-g-2-t-h-2-t-f-2-u-du-where-h-1-g-1-h-2-and-g-2-are-different

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Question Number 2284 by Yozzi last updated on 13/Nov/15
Define a curve E by the parametric equations x(t)=∫_(g_1 (t)) ^(h_1 (t)) f_1 (u)du y(t)=∫_(g_2 (t)) ^(h_2 (t)) f_2 (u)du where h_1 ,g_1 ,h_2 and g_2 are differentiable and both f_1 and f_2 are continuous. Find the gradient, m, of the tangent to E at the point N with parameter value t (t≥0). Let h_1 (t)=t,g_1 (t)=t^2 ,h_2 (t)=t^3 ,g_2 (t)=t^4 f_1 (u)=u^2 and f_2 (u)=u^3 . Find the values of t when the tangent is (i) vertical, (ii) horizontal. What are the equations of the tangents in (i) and (ii)?
$${Define}\:{a}\:{curve}\:{E}\:{by}\:{the}\:{parametric} \\ $$$${equations} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}\left({t}\right)=\int_{{g}_{\mathrm{1}} \left({t}\right)} ^{{h}_{\mathrm{1}} \left({t}\right)} {f}_{\mathrm{1}} \left({u}\right){du} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{y}\left({t}\right)=\int_{{g}_{\mathrm{2}} \left({t}\right)} ^{{h}_{\mathrm{2}} \left({t}\right)} {f}_{\mathrm{2}} \left({u}\right){du} \\ $$$${where}\:{h}_{\mathrm{1}} ,{g}_{\mathrm{1}} ,{h}_{\mathrm{2}} \:{and}\:{g}_{\mathrm{2}} \:{are}\:{differentiable} \\ $$$${and}\:{both}\:{f}_{\mathrm{1}} \:{and}\:{f}_{\mathrm{2}} \:{are}\:{continuous}. \\ $$$${Find}\:{the}\:{gradient},\:{m},\:{of}\:{the}\:{tangent} \\ $$$${to}\:{E}\:{at}\:{the}\:{point}\:{N}\:{with}\:{parameter}\: \\ $$$${value}\:{t}\:\left({t}\geqslant\mathrm{0}\right). \\ $$$${Let}\:{h}_{\mathrm{1}} \left({t}\right)={t},{g}_{\mathrm{1}} \left({t}\right)={t}^{\mathrm{2}} ,{h}_{\mathrm{2}} \left({t}\right)={t}^{\mathrm{3}} ,{g}_{\mathrm{2}} \left({t}\right)={t}^{\mathrm{4}} \\ $$$${f}_{\mathrm{1}} \left({u}\right)={u}^{\mathrm{2}} \:{and}\:{f}_{\mathrm{2}} \left({u}\right)={u}^{\mathrm{3}} .\:{Find}\:{the}\: \\ $$$${values}\:{of}\:{t}\:{when}\:{the}\:{tangent}\:{is}\: \\ $$$$\left({i}\right)\:{vertical}, \\ $$$$\left({ii}\right)\:{horizontal}. \\ $$$${What}\:{are}\:{the}\:{equations}\:{of}\:{the}\:{tangents} \\ $$$${in}\:\left({i}\right)\:{and}\:\left({ii}\right)? \\ $$$$ \\ $$
Commented by Yozzi last updated on 14/Nov/15
A consequence of the Fundamental Theorem of Calculus is that, for φ being continuous , and both ψ and η being differentiable, (d/dx)∫_(η(x)) ^(ψ(x)) φ(z)dz=φ(ψ(x))ψ^′ (x)−φ(η(x))η^′ (x). The Fundamental Theorem of Calculus states that for f being continuous over the interval a≤t≤b, with one given any x∈[a,b], and we let F be a function such that F(x)=∫_a ^x f(t)dt then F^( ′) (x)=f(x). It may interest you to prove the result using this theorem before using the result.
$${A}\:{consequence}\:{of}\:{the}\:{Fundamental} \\ $$$${Theorem}\:{of}\:{Calculus}\:{is}\:{that},\:{for}\: \\ $$$$\phi\:\:{being}\:{continuous}\:,\:{and}\:{both}\:\psi\:{and}\:\eta \\ $$$${being}\:{differentiable}, \\ $$$$\frac{{d}}{{dx}}\int_{\eta\left({x}\right)} ^{\psi\left({x}\right)} \phi\left({z}\right){dz}=\phi\left(\psi\left({x}\right)\right)\psi^{'} \left({x}\right)−\phi\left(\eta\left({x}\right)\right)\eta^{'} \left({x}\right). \\ $$$$ \\ $$$${The}\:{Fundamental}\:{Theorem}\:{of}\:{Calculus} \\ $$$${states}\:{that}\:{for}\:{f}\:{being}\:{continuous}\:{over} \\ $$$${the}\:{interval}\:{a}\leqslant{t}\leqslant{b},\:{with}\:{one}\:{given} \\ $$$${any}\:{x}\in\left[{a},{b}\right],\:{and}\:{we}\:{let}\:{F}\:{be}\:{a}\:{function} \\ $$$${such}\:{that} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{F}\left({x}\right)=\int_{{a}} ^{{x}} {f}\left({t}\right){dt} \\ $$$${then}\:{F}^{\:'} \left({x}\right)={f}\left({x}\right).\: \\ $$$${It}\:{may}\:{interest}\:{you}\:{to}\:{prove}\:{the}\:{result} \\ $$$${using}\:{this}\:{theorem}\:{before}\:{using}\:{the}\: \\ $$$${result}. \\ $$
Answered by Yozzi last updated on 15/Nov/15
We need to find x^′ (t) and y^′ (t). x^′ (t)=(d/dt)∫_(g_1 (t)) ^(h_1 (t)) f_1 (u)du. From comments, x^′ (t)=f_1 (h_1 (t))h_1 ^′ (t)−f_1 (g_1 (t))g_1 ^′ (t) Similarly, for y^′ (t)=(d/dt)∫_(g_2 (t)) ^(h_2 (t)) f_2 (u)du, y^′ (t)=f_2 (h_2 (t))h_2 ^′ (t)−f_2 (g_2 (t))g_2 ^′ (t). The gradient m of the tangent to E at the point N with parameter value t (t≥0) is then given by m=((y^′ (t))/(x^′ (t))). m=((f_2 (h_2 (t))h_2 ^′ (t)−f_2 (g_2 (t))g_2 ^′ (t))/(f_1 (h_1 (t))h_1 ^′ (t)−f_1 (g_1 (t))g_1 ^′ (t))). Given h_1 (t)=t,g_1 (t)=t^2 ,h_2 (t)=t^3 ,g_2 (t)=t^4 ⇒h_1 ^′ (t)=1, g_1 ^′ (t)=2t,h_2 ^′ (t)=3t^2 ,g_2 ^′ (t)=4t^3 . ∵ f_1 (u)=u^2 ,f_2 (u)=u^3 m=(((t^3 )^3 ×3t^2 −(t^4 )^3 ×4t^3 )/((t)^2 ×1−(t^2 )^2 ×2t)) =((3t^(11) −4t^(15) )/(t^2 −2t^5 )) =((t^(11) (3−4t^4 ))/(t^2 (1−2t^3 ))) m=((t^9 (3−4t^4 ))/(1−2t^3 )) (i) If the tangent to E at N is vertical ⇒m is undefined for some t. This occurs if 1−2t^3 =0⇒t=2^(−(1/3)) . (ii)If the tangent to E at N is horizontal, ⇒m=0⇒t^9 (3−4t^4 )=0⇒t^9 =0 or 3−4t^4 =0. t^9 =0⇒t=0. 3−4t^4 =0⇒t^4 =3/4⇒t=±((3/4))^(1/4) . Since t≥0 at N⇒t=((3/4))^(1/4) . t=0 ∨ t=((3/4))^(1/4) .
$${We}\:{need}\:{to}\:{find}\:{x}^{'} \left({t}\right)\:{and}\:{y}^{'} \left({t}\right).\: \\ $$$${x}^{'} \left({t}\right)=\frac{{d}}{{dt}}\int_{{g}_{\mathrm{1}} \left({t}\right)} ^{{h}_{\mathrm{1}} \left({t}\right)} {f}_{\mathrm{1}} \left({u}\right){du}.\:{From}\:{comments}, \\ $$$${x}^{'} \left({t}\right)={f}_{\mathrm{1}} \left({h}_{\mathrm{1}} \left({t}\right)\right){h}_{\mathrm{1}} ^{'} \left({t}\right)−{f}_{\mathrm{1}} \left({g}_{\mathrm{1}} \left({t}\right)\right){g}_{\mathrm{1}} ^{'} \left({t}\right) \\ $$$${Similarly},\:{for}\:{y}^{'} \left({t}\right)=\frac{{d}}{{dt}}\int_{{g}_{\mathrm{2}} \left({t}\right)} ^{{h}_{\mathrm{2}} \left({t}\right)} {f}_{\mathrm{2}} \left({u}\right){du}, \\ $$$${y}^{'} \left({t}\right)={f}_{\mathrm{2}} \left({h}_{\mathrm{2}} \left({t}\right)\right){h}_{\mathrm{2}} ^{'} \left({t}\right)−{f}_{\mathrm{2}} \left({g}_{\mathrm{2}} \left({t}\right)\right){g}_{\mathrm{2}} ^{'} \left({t}\right). \\ $$$${The}\:{gradient}\:{m}\:{of}\:{the}\:{tangent}\:{to}\:{E}\:{at}\:{the} \\ $$$${point}\:{N}\:{with}\:{parameter}\:{value}\:{t}\:\left({t}\geqslant\mathrm{0}\right) \\ $$$${is}\:{then}\:{given}\:{by} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{m}=\frac{{y}^{'} \left({t}\right)}{{x}^{'} \left({t}\right)}. \\ $$$$\:\:\:\:{m}=\frac{{f}_{\mathrm{2}} \left({h}_{\mathrm{2}} \left({t}\right)\right){h}_{\mathrm{2}} ^{'} \left({t}\right)−{f}_{\mathrm{2}} \left({g}_{\mathrm{2}} \left({t}\right)\right){g}_{\mathrm{2}} ^{'} \left({t}\right)}{{f}_{\mathrm{1}} \left({h}_{\mathrm{1}} \left({t}\right)\right){h}_{\mathrm{1}} ^{'} \left({t}\right)−{f}_{\mathrm{1}} \left({g}_{\mathrm{1}} \left({t}\right)\right){g}_{\mathrm{1}} ^{'} \left({t}\right)}. \\ $$$${Given}\:{h}_{\mathrm{1}} \left({t}\right)={t},{g}_{\mathrm{1}} \left({t}\right)={t}^{\mathrm{2}} ,{h}_{\mathrm{2}} \left({t}\right)={t}^{\mathrm{3}} ,{g}_{\mathrm{2}} \left({t}\right)={t}^{\mathrm{4}} \\ $$$$\Rightarrow{h}_{\mathrm{1}} ^{'} \left({t}\right)=\mathrm{1},\:{g}_{\mathrm{1}} ^{'} \left({t}\right)=\mathrm{2}{t},{h}_{\mathrm{2}} ^{'} \left({t}\right)=\mathrm{3}{t}^{\mathrm{2}} ,{g}_{\mathrm{2}} ^{'} \left({t}\right)=\mathrm{4}{t}^{\mathrm{3}} . \\ $$$$\because\:{f}_{\mathrm{1}} \left({u}\right)={u}^{\mathrm{2}} ,{f}_{\mathrm{2}} \left({u}\right)={u}^{\mathrm{3}} \\ $$$${m}=\frac{\left({t}^{\mathrm{3}} \right)^{\mathrm{3}} ×\mathrm{3}{t}^{\mathrm{2}} −\left({t}^{\mathrm{4}} \right)^{\mathrm{3}} ×\mathrm{4}{t}^{\mathrm{3}} }{\left({t}\right)^{\mathrm{2}} ×\mathrm{1}−\left({t}^{\mathrm{2}} \right)^{\mathrm{2}} ×\mathrm{2}{t}} \\ $$$$\:\:\:\:=\frac{\mathrm{3}{t}^{\mathrm{11}} −\mathrm{4}{t}^{\mathrm{15}} }{{t}^{\mathrm{2}} −\mathrm{2}{t}^{\mathrm{5}} } \\ $$$$\:\:\:\:=\frac{{t}^{\mathrm{11}} \left(\mathrm{3}−\mathrm{4}{t}^{\mathrm{4}} \right)}{{t}^{\mathrm{2}} \left(\mathrm{1}−\mathrm{2}{t}^{\mathrm{3}} \right)} \\ $$$${m}=\frac{{t}^{\mathrm{9}} \left(\mathrm{3}−\mathrm{4}{t}^{\mathrm{4}} \right)}{\mathrm{1}−\mathrm{2}{t}^{\mathrm{3}} } \\ $$$$\left({i}\right)\:{If}\:{the}\:{tangent}\:{to}\:{E}\:{at}\:{N}\:{is}\:{vertical} \\ $$$$\Rightarrow{m}\:{is}\:{undefined}\:{for}\:{some}\:{t}. \\ $$$${This}\:{occurs}\:{if}\:\mathrm{1}−\mathrm{2}{t}^{\mathrm{3}} =\mathrm{0}\Rightarrow{t}=\mathrm{2}^{−\frac{\mathrm{1}}{\mathrm{3}}} . \\ $$$$\left({ii}\right){If}\:{the}\:{tangent}\:{to}\:{E}\:{at}\:{N}\:{is}\:{horizontal}, \\ $$$$\Rightarrow{m}=\mathrm{0}\Rightarrow{t}^{\mathrm{9}} \left(\mathrm{3}−\mathrm{4}{t}^{\mathrm{4}} \right)=\mathrm{0}\Rightarrow{t}^{\mathrm{9}} =\mathrm{0}\:{or}\:\mathrm{3}−\mathrm{4}{t}^{\mathrm{4}} =\mathrm{0}. \\ $$$${t}^{\mathrm{9}} =\mathrm{0}\Rightarrow{t}=\mathrm{0}. \\ $$$$\mathrm{3}−\mathrm{4}{t}^{\mathrm{4}} =\mathrm{0}\Rightarrow{t}^{\mathrm{4}} =\mathrm{3}/\mathrm{4}\Rightarrow{t}=\pm\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{4}} . \\ $$$${Since}\:{t}\geqslant\mathrm{0}\:{at}\:{N}\Rightarrow{t}=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{4}} . \\ $$$${t}=\mathrm{0}\:\vee\:{t}=\left(\frac{\mathrm{3}}{\mathrm{4}}\right)^{\mathrm{1}/\mathrm{4}} . \\ $$
Answered by Yozzi last updated on 15/Nov/15
Given the functions of h_1 ,g_1 ,h_2 ,g_2 ,f_(1 ) and f_2 , we have x(t)=∫_t^2 ^( t) u^2 du=(u^3 /3)∣_t^2 ^t =((t^3 −t^6 )/3) y(t)=∫_t^4 ^( t^3 ) u^3 du=(u^4 /4)∣_t^4 ^t^3 =((t^(12) −t^(16) )/4). If the tangent is vertical at t=2^(−1/3) , we require an x−line. ∴x(2^(−1/3) )=((2^(−1) −2^(−2) )/3)=(1/(12)) ⇒ equation of vertical tangent is x=(1/(12)). If the tangnet is horizontal we have a y−line. For t=0,y(0)=0. ∴ the equation of the tangent is y=0. For t=(3/4)^(1/4) , y((3/4)^(1/4) )=((0.75^3 −0.75^4 )/4)=((((27)/(64))−((81)/(256)))/4) y((3/4)^(1/4) )=(1/4)×((108−81)/(256))=((27)/(1024)) The equation of the other horizontal tangent is y=27/1024.
$${Given}\:{the}\:{functions}\:{of}\:{h}_{\mathrm{1}} ,{g}_{\mathrm{1}} ,{h}_{\mathrm{2}} ,{g}_{\mathrm{2}} ,{f}_{\mathrm{1}\:} \\ $$$${and}\:{f}_{\mathrm{2}} ,\:{we}\:{have} \\ $$$${x}\left({t}\right)=\int_{{t}^{\mathrm{2}} } ^{\:{t}} {u}^{\mathrm{2}} {du}=\frac{{u}^{\mathrm{3}} }{\mathrm{3}}\mid_{{t}^{\mathrm{2}} } ^{{t}} =\frac{{t}^{\mathrm{3}} −{t}^{\mathrm{6}} }{\mathrm{3}} \\ $$$${y}\left({t}\right)=\int_{{t}^{\mathrm{4}} } ^{\:{t}^{\mathrm{3}} } {u}^{\mathrm{3}} {du}=\frac{{u}^{\mathrm{4}} }{\mathrm{4}}\mid_{{t}^{\mathrm{4}} } ^{{t}^{\mathrm{3}} } =\frac{{t}^{\mathrm{12}} −{t}^{\mathrm{16}} }{\mathrm{4}}. \\ $$$${If}\:{the}\:{tangent}\:{is}\:{vertical}\:{at}\:{t}=\mathrm{2}^{−\mathrm{1}/\mathrm{3}} , \\ $$$${we}\:{require}\:{an}\:{x}−{line}.\: \\ $$$$\therefore{x}\left(\mathrm{2}^{−\mathrm{1}/\mathrm{3}} \right)=\frac{\mathrm{2}^{−\mathrm{1}} −\mathrm{2}^{−\mathrm{2}} }{\mathrm{3}}=\frac{\mathrm{1}}{\mathrm{12}} \\ $$$$\Rightarrow\:{equation}\:{of}\:{vertical}\:{tangent}\:{is}\:{x}=\frac{\mathrm{1}}{\mathrm{12}}. \\ $$$${If}\:{the}\:{tangnet}\:{is}\:{horizontal}\:{we}\:{have} \\ $$$${a}\:{y}−{line}.\:{For}\:{t}=\mathrm{0},{y}\left(\mathrm{0}\right)=\mathrm{0}. \\ $$$$\therefore\:{the}\:{equation}\:{of}\:{the}\:{tangent}\:{is}\:{y}=\mathrm{0}. \\ $$$${For}\:{t}=\left(\mathrm{3}/\mathrm{4}\right)^{\mathrm{1}/\mathrm{4}} , \\ $$$${y}\left(\left(\mathrm{3}/\mathrm{4}\right)^{\mathrm{1}/\mathrm{4}} \right)=\frac{\mathrm{0}.\mathrm{75}^{\mathrm{3}} −\mathrm{0}.\mathrm{75}^{\mathrm{4}} }{\mathrm{4}}=\frac{\frac{\mathrm{27}}{\mathrm{64}}−\frac{\mathrm{81}}{\mathrm{256}}}{\mathrm{4}} \\ $$$${y}\left(\left(\mathrm{3}/\mathrm{4}\right)^{\mathrm{1}/\mathrm{4}} \right)=\frac{\mathrm{1}}{\mathrm{4}}×\frac{\mathrm{108}−\mathrm{81}}{\mathrm{256}}=\frac{\mathrm{27}}{\mathrm{1024}} \\ $$$${The}\:{equation}\:{of}\:{the}\:{other}\:{horizontal} \\ $$$${tangent}\:{is}\:{y}=\mathrm{27}/\mathrm{1024}. \\ $$

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