matematikk.net

hey, er veldig usikker på denne oppgaven.. håper jeg kan få litt veiledning:)

find and equation for the line tangent to the curve at the point defined by the given value of t. Also,find the value of [tex]\frac{d^2 y}{dx^2}[/tex].

x=t-sint
y=1-cost
t=[tex]\frac{\pi}{3}[/tex]

Punktet [tex] (x,y) = (\frac{\pi}{3} -sin(\frac{\pi}{3}) , 1-cos(\frac{\pi}{3})) = (\frac{\pi}{3}-\frac{\sqrt{3}}{2} , 1-\frac{1}{2})=(\frac{\pi}{3}-\frac{\sqrt{3}}{2} , \frac{1}{2})[/tex]

Er ikke noe fasit i boken, men skal de fram til at vi skal gjøre noe slikt?

Stigningstall til tangenten:

[tex]y(u) = 1-cosu [/tex]
[tex]y(x) = 1-cos(x-sinx)[/tex]

[tex]\frac{d}{dx} (1-cos(x-sinx)) = sin(x-sinx)*(1-cosx)[/tex]

ah

[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}}[/tex]

[tex]\frac{dy}{dt}=\frac{d}{dt} (1-cost)=\frac{d}{dt} (1) + \frac{d}{dt} (-cost) = sint [/tex]

[tex]\frac{dx}{dt}=\frac{d}{dt} (t-sint)=\frac{d}{dt} (t) + \frac{d}{dt} (-sint)= 1-cost[/tex]

[tex]\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{sin(\frac{\pi}{3})}{1-cos(\frac{\pi}{3})}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}} = \sqrt{3}[/tex]

Tangent:

[tex]T(x)=\sqrt{3}(x-(\frac{\pi}{3}-\frac{\sqrt{3}}{2}))+\frac{1}{2}[/tex]

[tex]T(x)=\sqrt{3}(x-\frac{\pi}{3}+\frac{\sqrt{3}}{2})+\frac{1}{2}[/tex]
[tex]T(x)=\sqrt{3}x-\frac{\sqrt{3}\pi}{3}+\frac{3}{2}+\frac{1}{2}[/tex]
[tex]T(x)=\sqrt{3}x-\frac{\sqrt{3}\pi}{3}+2[/tex]

Ble ikke dette utrykket litt styggt?