16 KiB
excalidraw-plugin, tags
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==⚠ Switch to EXCALIDRAW VIEW in the MORE OPTIONS menu of this document. ⚠== You can decompress Drawing data with the command palette: 'Decompress current Excalidraw file'. For more info check in plugin settings under 'Saving'
Excalidraw Data
Text Elements
le tableau de variations est évident étant donné le problème considéré. On pourra toujours montrer que cet unique extrémum est bien un maximum à partir du fait que 𝑎 ∈ [0; π/2] ^OnHRRJcd
Embedded Files
8923efd922b888d467179262d7b788641d043cc7: l
b3d5701da7f512bb6c2cfeefa687ef9448961a27: L
a745f1dc9892805697eb7dc6df5134ce60d3e778: \alpha
b76ee720d7e0bdf63f408ee39db94f3d31ca9f0e: r
13099823a77cd5a2252736e0f029c959c6ee8e2c: =\begin{pmatrix}l\\L\end{pmatrix}
e8b51e7e123fde3009046ff7524c05292519716e: =\begin{pmatrix}r \cos \alpha\\ r \sin \alpha\end{pmatrix}
2eb73fd7a8bd07b84080b3bb149f4819930587be: = \begin{pmatrix}l + r \cos \alpha\\ L + r \sin \alpha\end{pmatrix}
4dd7657e8aff3434304e0c34035d8f6e3e9c7c5d: \text{donc :}
ce7a3ec53f89ae18583ec0042233662e411ba23b: \begin{align} \left\| \begin{pmatrix}l+r\cos \alpha\\L+r \sin \alpha\end{pmatrix} \right\| &= l^{2}+2lr\cos\alpha + r^{2}\cos ^{2}\alpha + L^{2} + 2Lr\sin\alpha + r^{2}\sin ^{2}\alpha \\&= (l^{2}+L^{2}+r^{2}) + 2r(l\cos\alpha+L\sin\alpha)\end{align}
16647f064fb7a498cb3e96cfa483672e387b3557: \text{on veut donc maximiser } f: f(\alpha) = l\cos\alpha+L\sin\alpha
c07e13d38824e5b0021dd5fdfadb2c55ad79c69a: f'(\alpha) = -l\sin\alpha + L\cos\alpha
fafa660ca9b7e22e4de89da24218f9436103d840: \begin{align} f'(\alpha) = 0 &\iff L \cos \alpha = l \sin \alpha \\&\iff \frac{L}{l} \frac{\cos \alpha}{\sin\alpha} = 1 \\&\iff \tan \alpha = \frac{l}{L} \\&\iff \boxed {\alpha = \arctan \frac{l}{L}} \end{align}
%%
Drawing
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