Question

Find a vector normal to the surface $x^2+y^2-z=0$ at the point (3,4,25) . Find the equations of the tangent plane and normal line to the surface at that point.

Short answer

Gradient vector at (3,4,25) is (6,8,-1). Tangent plane equation: 6x+8y-z=25. Normal line equations: x=3+6t, y=4+8t, z=25-t.

Step-by-step solution

Find the gradient vector

Given the surface equation is $$x^2+y^2-z=0$$ Calculate the gradient vector $ablaF$ of $F(x,y,z)=x^2+y^2-z$. The gradient vector $ablaF=(\frac{\partial F}{\partial x},\frac{\partial F}{\partial y},\frac{\partial F}{\partial z})$.

Compute partial derivatives

Find the partial derivatives of $F(x,y,z)$: $$\frac{\partial F}{\partial x}=2x$$ $$\frac{\partial F}{\partial y}=2y$$ $$\frac{\partial F}{\partial z}=-1$$

Evaluate the gradient at the given point

Substitute $(x,y,z)=(3,4,25)$ into the gradient: $$ablaF(3,4,25)=(2\cdot 3,2\cdot 4,-1)=(6,8,-1)$$. This is the normal vector to the surface at the point (3,4,25).

Find the equation of the tangent plane

The equation of the tangent plane at the point $(x_0,y_0,z_0)$ is: $$(ablaF)\cdot ((x-x_0),(y-y_0),(z-z_0))=0$$ Using the normal vector $ablaF(3,4,25)=(6,8,-1)$,and the point (3,4,25), the tangent plane equation is: $$6(x-3)+8(y-4)-(z-25)=0$$ Simplifying:$$6x+8y-z=25$$

Find the parametric equations of the normal line

The normal line through the point $(3,4,25)$ with direction ratios $(6,8,-1)$: The parametric equations of the normal line are: $$x=3+6t$$ $$y=4+8t$$ $$z=25-t$$

Key concepts

Normal Vector

In vector calculus, the normal vector to a surface at a given point is a crucial concept. It is perpendicular to the tangent plane of the surface at that point. For the surface described by the equation $x^2+y^2-z=0$, the gradient vector at any point gives the normal vector to the surface at that point. To find this, compute the partial derivatives of the function $F(x,y,z)=x^2+y^2-z$. This yields the components of the normal vector as follows:

• $\frac{\partial F}{\partial x}=2x$
• $\frac{\partial F}{\partial y}=2y$
• $\frac{\partial F}{\partial z}=-1$

At the point (3, 4, 25), substituting the values into the partial derivatives, we get the normal vector $(6,8,-1)$. This vector is perpendicular to the surface at the given point.

Gradient Vector

The gradient vector is an essential tool in vector calculus. It is a vector that points in the direction of the greatest rate of increase of the function. For the function $F(x,y,z)=x^2+y^2-z$, the gradient vector $ablaF$ is calculated by taking the partial derivatives of F with respect to x, y, and z. This provides the components of the gradient vector:

• $\frac{\partial F}{\partial x}=2x$
• $\frac{\partial F}{\partial y}=2y$
• $\frac{\partial F}{\partial z}=-1$

By substituting the coordinates of the point (3, 4, 25) into these partial derivatives, we get the gradient vector $(6,8,-1)$. This vector is not only the gradient but also the normal vector to the surface at that point.

Tangent Plane

The tangent plane to a surface at a point provides a linear approximation of the surface around that point. To determine the equation of the tangent plane for the surface $x^2+y^2-z=0$ at the point (3, 4, 25), use the gradient vector $(6,8,-1)$. The equation of the tangent plane is given by:$$(ablaF)\bullet (x-x_0,y-y_0,z-z_0)=0$$Substituting $ablaF=(6,8,-1)$\ and the point $(3,4,25)$, we get:$$6(x-3)+8(y-4)-(z-25)=0$$Simplify this to:$$6x+8y-z=25$$This is the equation of the tangent plane at the given point.

Normal Line

The normal line to a surface at a given point is a line that passes through the point and is perpendicular to the tangent plane. For the surface $x^2+y^2-z=0$ at the point (3, 4, 25), the normal line has direction ratios given by the components of the normal vector $(6,8,-1)$. The parametric equations of the normal line can be written as:

• $x=3+6t$
• $y=4+8t$
• $z=25-t$

These equations describe the normal line that passes through the point $(3,4,25)$ and follows the direction of the normal vector.

Partial Derivatives

Partial derivatives represent the rate of change of a function with respect to one variable while keeping other variables constant. For the function $F(x,y,z)=x^2+y^2-z$, the partial derivatives with respect to x, y, and z are:

• $\frac{\partial F}{\partial x}=2x$
• $\frac{\partial F}{\partial y}=2y$
• $\frac{\partial F}{\partial z}=-1$

By substituting the coordinates $(3,4,25)$ into these expressions, we find the values of these partial derivatives, yielding the components of the gradient vector:$(6,8,-1)$. Partial derivatives are essential in calculating the gradient vector and, consequently, the normal vector to the surface at a given point.