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Flashcards: Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(0,5) $\displaystyle z=(0,5)$.

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$ $\displaystyle z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

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Flashcards: Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(0,5) $\displaystyle z=(0,5)$.

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$ $\displaystyle z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

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Flashcards: Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(0,5) $\displaystyle z=(0,5)$.

Your answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

Your answer:

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

Your answer:

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

Your answer:

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$ $\displaystyle z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to $\displaystyle x$, and $\displaystyle y$, and plug in z=(1,5) $\displaystyle z=(1,5)$.

$f(x,y)=\ln(4x^3+10y^2)$ $\displaystyle f(x,y)=\ln(4x^3+10y^2)$, $f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$ $\displaystyle f(0,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ $\displaystyle f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$, $f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$ $\displaystyle f_x(0,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ $\displaystyle f_y(x,y)=\frac{20y}{4x^3+10y^2}$, $f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$ $\displaystyle f_y(0,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

$\displaystyle z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$ $\displaystyle z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$ $\displaystyle z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Flashcards: Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(1,5) $\displaystyle z=(1,5)$.

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$ $\displaystyle z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

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Flashcards: Matrices & Vectors

Find the equation of the tangent plane to $z=\ln(4x^3+10y^2)$ $\displaystyle z=\ln(4x^3+10y^2)$ at z=(1,5) $\displaystyle z=(1,5)$.

Your answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}$

Your answer:

$z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{50}{127}y+\frac{256}{127}+\ln(254)$

Your answer:

$z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y+\ln(254)$

Your answer:

$z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$ $\displaystyle z=-\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}-\ln(254)$

Correct answer:

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

Explanation:

First, we need to find the partial derivatives in respect to $\displaystyle x$, and $\displaystyle y$, and plug in z=(1,5) $\displaystyle z=(1,5)$.

$f(x,y)=\ln(4x^3+10y^2)$ $\displaystyle f(x,y)=\ln(4x^3+10y^2)$, $f(1,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$ $\displaystyle f(1,5)=\ln(4(1)^3+10(5)^2)=\ln(254)$

$f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$ $\displaystyle f_x(x,y)=\frac{12x^2}{4x^3+10y^2}$, $f_x(1,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$ $\displaystyle f_x(1,5)=\frac{12(1)^2}{4(1)^3+10(5)^2}=\frac{12}{254}=\frac{6}{127}$

$f_y(x,y)=\frac{20y}{4x^3+10y^2}$ $\displaystyle f_y(x,y)=\frac{20y}{4x^3+10y^2}$, $f_y(1,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$ $\displaystyle f_y(1,5)=\frac{20(5)}{4(1)^3+10(5)^2}=\frac{100}{254}=\frac{50}{127}$

Remember that the general equation for a tangent plane is as follows:

$\displaystyle z-f(x_0,y_0)=f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$

Now lets apply this to our problem

$z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$ $\displaystyle z-\ln(254)=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)$

$z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$ $\displaystyle z=\frac{6}{127}(x-1)+\frac{50}{127}(y-5)+\ln(254)$

$z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x-\frac{6}{127}+\frac{50}{127}y-\frac{250}{127}+\ln(254)$

$z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$ $\displaystyle z=\frac{6}{127}x+\frac{50}{127}y-\frac{256}{127}+\ln(254)$

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Flashcards: Matrices & Vectors

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