Looking ahead- tangent planes Consider the following...

Looking ahead- tangent planes Consider the following surfaces $f(x, y, z)=0,$ which may be regarded as a level surface of the function $w=f(x, y, z) .$ A point $P(a, b, c)$ on the surface is also given. a. Find the (three-dimensional) gradient of $f$ and evaluate it at $P$. b. The set of all vectors orthogonal to the gradient with their tails at $P$ form a plane. Find an equation of that plane (soon to be called the tangent plane). $$f(x, y, z)=8-x y z=0 ; P(2,2,2)$$

Looking ahead- tangent planes Consider the following surfaces $f(x, y, z)=0,$ which may be regarded as a level surface of the
function $w=f(x, y, z) .$ A point $P(a, b, c)$ on the surface is also given.
a. Find the (three-dimensional) gradient of $f$ and evaluate it at $P$.
b. The set of all vectors orthogonal to the gradient with their tails at $P$ form a plane. Find an equation of that plane (soon to be called the tangent plane).
$$f(x, y, z)=8-x y z=0 ; P(2,2,2)$$

Key Concepts

Level Surfaces

A level surface is defined by an equation of the form f(x, y, z) = c, where every point (x, y, z) on the surface satisfies the equation. It represents the set of points in space where the function f has the same constant value, providing a geometric way to analyze the properties of multivariable functions.

Gradient Vector

The gradient vector, denoted as ?f, is a vector that consists of the partial derivatives of a function with respect to its variables. It points in the direction of the greatest increase of the function and its magnitude represents the rate of change in that direction. Importantly, for level surfaces, the gradient is perpendicular to the surface, making it a key tool for finding normal vectors.

Tangent Plane

The tangent plane to a surface at a particular point is the plane that best approximates the surface near that point. It is defined using the gradient vector, since the gradient is normal to the surface. The plane consists of all directions perpendicular to the gradient vector, and its equation can be derived using the point-normal form of a plane.

Partial Derivatives

Partial derivatives are the derivatives of a function with respect to one variable while keeping the other variables fixed. They are the building blocks of the gradient vector and are essential for determining the rate and direction of change of the function with multiple variables.

Transcript

00:04 In this problem, we have to find the gradient of the given function f and then evaluate it at the given point p.

00:17 Gradient of f, that is del f, is equal to the first component of del f is f is fx, that is partial derivative of f with respect to x.

00:37 The next one is f y that is partial derivative of f with respect to y and the last word is f z that is partial derivative of f with respect to z so this is equal to partial derivative of the given function f with respect to x is minus y z to y is minus x z and partial derivative of f with respect to z is minus x y so this is the gradient of the given function f now we have to evaluate it at the given point p two to two two so we have to evaluate del f at 2 at 2 2 2 2 that is we put the value of x, y and z, all three equal to two.

01:57 So minus yz, that is minus 2, multiplied by 2 is minus 4.

02:06 Similarly, minus x, z is also 4 and minus xy is also 4.

02:17 That is del f at 2 to 2 is equal to minus 4 minus 4 minus 4 minus 4 in the next subpart it is asked to find the equation of the tangent plane at the same point p 2 to 2 2 2 .2.

02:40 Equation of tangent plane at point 2 to 2 is given by f x at 2 to 2 that is partial derivative of f with respect to x at the given point multiplied by x minus x coordinate of the given point that is 2 plus f y at 2 2 2 that is the partial derivative of the given function with respect to y at the given point multiplied by y minus y coordinate of the given point which is 2 plus f z at 2 2 2 that is partial derivative of given function with respect to z at the given point 2 to 2 multiplied by z minus z coordinate of the given point which is 2 is 2 is equal to 0 so, fx at 2 to 2 is minus 4...

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