Tangents & Normals - Multivariable Calculus

Card 0 of 8

Find the equation of the tangent plane to at .

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First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$

Find the equation of the tangent plane to at .

Tap to see back →

First, we need to find the partial derivatives in respect to , and , and plug in .

,

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

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

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

Now lets apply this to our problem

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

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

$z=\frac{6}{127}$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}$+\ln(254)$