Ayuda con este ejercicio por favor.
Si la derivada direccional de F: R² --> R en (2,3) es 2 en la dirección que forma un angulo de 30° con el eje x positivo y 8 cuando este angulo es 150°. Determine cual es la derivada direccional de F en (2,3) en la dirección tangente a la curva definida por: G(x,y) = 2xy-3y² = 1 en (2,1)
Respuestas
Respuesta dada por:
2
La derivada direccional de F en dirección del vector unitario
se calcula así
![D_{\vec u}F=\nabla F\cdot \vec u D_{\vec u}F=\nabla F\cdot \vec u](https://tex.z-dn.net/?f=D_%7B%5Cvec+u%7DF%3D%5Cnabla+F%5Ccdot+%5Cvec+u)
1)![\vec u_1 = (\cos 30\°,\sin30\°)=\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right) \vec u_1 = (\cos 30\°,\sin30\°)=\left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)](https://tex.z-dn.net/?f=%5Cvec+u_1+%3D+%28%5Ccos+30%5C%C2%B0%2C%5Csin30%5C%C2%B0%29%3D%5Cleft%28%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%5Cright%29)
![D_{\vec u_1}F=\nabla F\cdot \vec u_1\\ \\
D_{\vec u_1}F=(F_x,F_y)\cdot \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\\ \\ \\
D_{\vec u_1}F=\dfrac{\sqrt{3}F_x}{2}+\dfrac{F_y}{2}\\ \\ \\
D_{\vec u_1}F(2,3)=\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}\\ \\ \\
\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}=2\\ \\
\sqrt{3}F_x(2,3)+F_y(2,3)=4~~~~\cdots\cdots\cdots\textcircled{1}
D_{\vec u_1}F=\nabla F\cdot \vec u_1\\ \\
D_{\vec u_1}F=(F_x,F_y)\cdot \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\\ \\ \\
D_{\vec u_1}F=\dfrac{\sqrt{3}F_x}{2}+\dfrac{F_y}{2}\\ \\ \\
D_{\vec u_1}F(2,3)=\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}\\ \\ \\
\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}=2\\ \\
\sqrt{3}F_x(2,3)+F_y(2,3)=4~~~~\cdots\cdots\cdots\textcircled{1}](https://tex.z-dn.net/?f=D_%7B%5Cvec+u_1%7DF%3D%5Cnabla+F%5Ccdot+%5Cvec+u_1%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_1%7DF%3D%28F_x%2CF_y%29%5Ccdot+%5Cleft%28%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_1%7DF%3D%5Cdfrac%7B%5Csqrt%7B3%7DF_x%7D%7B2%7D%2B%5Cdfrac%7BF_y%7D%7B2%7D%5C%5C+%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_1%7DF%282%2C3%29%3D%5Cdfrac%7B%5Csqrt%7B3%7DF_x%282%2C3%29%7D%7B2%7D%2B%5Cdfrac%7BF_y%282%2C3%29%7D%7B2%7D%5C%5C+%5C%5C+%5C%5C%0A%5Cdfrac%7B%5Csqrt%7B3%7DF_x%282%2C3%29%7D%7B2%7D%2B%5Cdfrac%7BF_y%282%2C3%29%7D%7B2%7D%3D2%5C%5C+%5C%5C%0A%5Csqrt%7B3%7DF_x%282%2C3%29%2BF_y%282%2C3%29%3D4%7E%7E%7E%7E%5Ccdots%5Ccdots%5Ccdots%5Ctextcircled%7B1%7D%0A)
2)![\vec u_2=(\cos 150,\sin 150)=\left(-\dfrac{\sqrt3}{2},\dfrac{1}{2}\right)
\vec u_2=(\cos 150,\sin 150)=\left(-\dfrac{\sqrt3}{2},\dfrac{1}{2}\right)](https://tex.z-dn.net/?f=%5Cvec+u_2%3D%28%5Ccos+150%2C%5Csin+150%29%3D%5Cleft%28-%5Cdfrac%7B%5Csqrt3%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%0A)
![D_{\vec u_2}F=\nabla F\cdot \vec u_2\\ \\
D_{\vec u_2}F=(F_x,F_y)\cdot \left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\\ \\ \\
D_{\vec u_2}F=-\dfrac{\sqrt{3}F_x}{2}+\dfrac{F_y}{2}\\ \\ \\
D_{\vec u_2}F(2,3)=-\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}\\ \\ \\
-\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}=8\\ \\
-\sqrt{3}F_x(2,3)+F_y(2,3)=16~~~~\cdots\cdots\cdots\textcircled{2}
D_{\vec u_2}F=\nabla F\cdot \vec u_2\\ \\
D_{\vec u_2}F=(F_x,F_y)\cdot \left(-\dfrac{\sqrt{3}}{2},\dfrac{1}{2}\right)\\ \\ \\
D_{\vec u_2}F=-\dfrac{\sqrt{3}F_x}{2}+\dfrac{F_y}{2}\\ \\ \\
D_{\vec u_2}F(2,3)=-\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}\\ \\ \\
-\dfrac{\sqrt{3}F_x(2,3)}{2}+\dfrac{F_y(2,3)}{2}=8\\ \\
-\sqrt{3}F_x(2,3)+F_y(2,3)=16~~~~\cdots\cdots\cdots\textcircled{2}](https://tex.z-dn.net/?f=D_%7B%5Cvec+u_2%7DF%3D%5Cnabla+F%5Ccdot+%5Cvec+u_2%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_2%7DF%3D%28F_x%2CF_y%29%5Ccdot+%5Cleft%28-%5Cdfrac%7B%5Csqrt%7B3%7D%7D%7B2%7D%2C%5Cdfrac%7B1%7D%7B2%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_2%7DF%3D-%5Cdfrac%7B%5Csqrt%7B3%7DF_x%7D%7B2%7D%2B%5Cdfrac%7BF_y%7D%7B2%7D%5C%5C+%5C%5C+%5C%5C%0AD_%7B%5Cvec+u_2%7DF%282%2C3%29%3D-%5Cdfrac%7B%5Csqrt%7B3%7DF_x%282%2C3%29%7D%7B2%7D%2B%5Cdfrac%7BF_y%282%2C3%29%7D%7B2%7D%5C%5C+%5C%5C+%5C%5C%0A-%5Cdfrac%7B%5Csqrt%7B3%7DF_x%282%2C3%29%7D%7B2%7D%2B%5Cdfrac%7BF_y%282%2C3%29%7D%7B2%7D%3D8%5C%5C+%5C%5C%0A-%5Csqrt%7B3%7DF_x%282%2C3%29%2BF_y%282%2C3%29%3D16%7E%7E%7E%7E%5Ccdots%5Ccdots%5Ccdots%5Ctextcircled%7B2%7D%0A)
(3) resolviendo el sistema
![\begin{cases}
\sqrt{3}F_x(2,3)+F_y(2,3)=4\\
-\sqrt{3}F_x(2,3)+F_y(2,3)=16
\end{cases}\\ \\ \\
F_x(2,3)=-2\sqrt{3}\\
F_y(2,3)=10\\ \\
\boxed{\nabla F(2,3)=\left(-2\sqrt{3},10\right)} \begin{cases}
\sqrt{3}F_x(2,3)+F_y(2,3)=4\\
-\sqrt{3}F_x(2,3)+F_y(2,3)=16
\end{cases}\\ \\ \\
F_x(2,3)=-2\sqrt{3}\\
F_y(2,3)=10\\ \\
\boxed{\nabla F(2,3)=\left(-2\sqrt{3},10\right)}](https://tex.z-dn.net/?f=%5Cbegin%7Bcases%7D%0A%5Csqrt%7B3%7DF_x%282%2C3%29%2BF_y%282%2C3%29%3D4%5C%5C%0A-%5Csqrt%7B3%7DF_x%282%2C3%29%2BF_y%282%2C3%29%3D16%0A%5Cend%7Bcases%7D%5C%5C+%5C%5C+%5C%5C%0AF_x%282%2C3%29%3D-2%5Csqrt%7B3%7D%5C%5C+%0AF_y%282%2C3%29%3D10%5C%5C+%5C%5C%0A%5Cboxed%7B%5Cnabla+F%282%2C3%29%3D%5Cleft%28-2%5Csqrt%7B3%7D%2C10%5Cright%29%7D)
(4) Hallemos el vector tangente a la curva generada por G
![\vec v =\nabla G \\ \\
\vec v =(G_x,G_y)\\ \\
\vec v = (2y,2x-6y)\\ \\
\text{En el punto }(2,1)\text{ tenemos}\\ \\
\vec v=(2,-2)\\ \\
\text{Ahora el vector unitario de }\vec v\\ \\
\vec u=\left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right) \vec v =\nabla G \\ \\
\vec v =(G_x,G_y)\\ \\
\vec v = (2y,2x-6y)\\ \\
\text{En el punto }(2,1)\text{ tenemos}\\ \\
\vec v=(2,-2)\\ \\
\text{Ahora el vector unitario de }\vec v\\ \\
\vec u=\left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)](https://tex.z-dn.net/?f=%5Cvec+v+%3D%5Cnabla+G+%5C%5C+%5C%5C%0A%5Cvec+v+%3D%28G_x%2CG_y%29%5C%5C+%5C%5C%0A%5Cvec+v+%3D+%282y%2C2x-6y%29%5C%5C+%5C%5C%0A%5Ctext%7BEn+el+punto+%7D%282%2C1%29%5Ctext%7B+tenemos%7D%5C%5C+%5C%5C%0A%5Cvec+v%3D%282%2C-2%29%5C%5C+%5C%5C%0A%5Ctext%7BAhora+el+vector+unitario+de+%7D%5Cvec+v%5C%5C+%5C%5C%0A%5Cvec+u%3D%5Cleft%28%5Cdfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2C-%5Cdfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Cright%29)
(5) Nos piden calcular![D_{\vec u}F(2,3)=\nabla F(2,3)\cdot \vec u D_{\vec u}F(2,3)=\nabla F(2,3)\cdot \vec u](https://tex.z-dn.net/?f=D_%7B%5Cvec+u%7DF%282%2C3%29%3D%5Cnabla+F%282%2C3%29%5Ccdot+%5Cvec+u)
![D_{\vec u}F(2,3)=\left(-2\sqrt{3},10\right)\cdot \left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)\\ \\ \\
D_{\vec u}F(2,3)=-\dfrac{2\sqrt{3}}{\sqrt{2}}-\dfrac{10}{\sqrt{2}}\\ \\ \\
\boxed{D_{\vec u}F(2,3)=-\dfrac{2\sqrt{3}+10}{\sqrt{2}}} D_{\vec u}F(2,3)=\left(-2\sqrt{3},10\right)\cdot \left(\dfrac{1}{\sqrt{2}},-\dfrac{1}{\sqrt{2}}\right)\\ \\ \\
D_{\vec u}F(2,3)=-\dfrac{2\sqrt{3}}{\sqrt{2}}-\dfrac{10}{\sqrt{2}}\\ \\ \\
\boxed{D_{\vec u}F(2,3)=-\dfrac{2\sqrt{3}+10}{\sqrt{2}}}](https://tex.z-dn.net/?f=D_%7B%5Cvec+u%7DF%282%2C3%29%3D%5Cleft%28-2%5Csqrt%7B3%7D%2C10%5Cright%29%5Ccdot+%5Cleft%28%5Cdfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%2C-%5Cdfrac%7B1%7D%7B%5Csqrt%7B2%7D%7D%5Cright%29%5C%5C+%5C%5C+%5C%5C%0AD_%7B%5Cvec+u%7DF%282%2C3%29%3D-%5Cdfrac%7B2%5Csqrt%7B3%7D%7D%7B%5Csqrt%7B2%7D%7D-%5Cdfrac%7B10%7D%7B%5Csqrt%7B2%7D%7D%5C%5C+%5C%5C+%5C%5C%0A%5Cboxed%7BD_%7B%5Cvec+u%7DF%282%2C3%29%3D-%5Cdfrac%7B2%5Csqrt%7B3%7D%2B10%7D%7B%5Csqrt%7B2%7D%7D%7D)
1)
2)
(3) resolviendo el sistema
(4) Hallemos el vector tangente a la curva generada por G
(5) Nos piden calcular
fernandomrb:
Muchisimas gracias
Preguntas similares
hace 6 años
hace 6 años
hace 6 años
hace 9 años
hace 9 años
hace 9 años
hace 9 años