Problem 15
Question
Find an equation of the plane tangent to the following surfaces at the given points. $$z^{2}-x^{2} / 16-y^{2} / 9-1=0 ;(4,3,-\sqrt{3}) \text { and }(-8,9, \sqrt{14})$$
Step-by-Step Solution
Verified Answer
Question: Find the equations of the tangent planes to the surface defined by the equation $$z^{2}-\frac{x^{2}}{16}-\frac{y^{2}}{9}-1=0$$ at the points \((4,3,-\sqrt{3})\) and \((-8,9, \sqrt{14})\).
Answer: The equations of the tangent planes at the given points are:
$$-\frac{1}{2}x-\frac{2}{3}y-2\sqrt{3}z-\frac{749}{36}=0$$
$$x-2y+2\sqrt{14}z-26=0$$
1Step 1: Find the gradient vector of the given surface
To find the gradient vector, we will differentiate the given surface equation with respect to x, y, and z. The given equation is
$$z^{2}-\frac{x^{2}}{16}-\frac{y^{2}}{9}-1=0$$
Taking partial derivatives with respect to x, y, and z, we have:
$$\nabla f(x,y,z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle = \left\langle -\frac{x}{8}, -\frac{2y}{9}, 2z \right\rangle$$
2Step 2: Find the gradient vector at the given points
Now, we will find the gradient vector at the given points \((4,3,-\sqrt{3})\) and \((-8,9, \sqrt{14})\).
For \((4,3,-\sqrt{3})\), plug the point into the gradient vector:
$$\nabla f(4,3,-\sqrt{3}) = \left\langle -\frac{1}{2}, -\frac{2}{3}, -2\sqrt{3} \right\rangle$$
For \((-8,9, \sqrt{14})\), plug the point into the gradient vector:
$$\nabla f(-8,9, \sqrt{14}) = \left\langle 1, -2, 2\sqrt{14} \right\rangle$$
3Step 3: Find the equation of the tangent plane at each point
Use the point-slope form for the equation of the tangent plane, which is:
$$f_x(x-x_0)+f_y(y-y_0)+f_z(z-z_0)=0$$
For \((4,3,-\sqrt{3})\), plug the point and gradient vector into the equation:
$$-\frac{1}{2}(x-4)-\frac{2}{3}(y-3)-2\sqrt{3}(z+\sqrt{3})=0$$
Simplify:
$$-\frac{1}{2}x+\frac{1}{4}-\frac{2}{3}y+\frac{1}{9}-2\sqrt{3}z-6=0$$
Combine constants:
$$-\frac{1}{2}x-\frac{2}{3}y-2\sqrt{3}z-\frac{749}{36}=0$$
For \((-8,9, \sqrt{14})\), plug the point and gradient vector into the equation:
$$1(x+8)-2(y-9)+2\sqrt{14}(z-\sqrt{14})=0$$
Simplify:
$$x+8-2y+18+2\sqrt{14}z-28\sqrt{14}=0$$
Combine constants:
$$x-2y+2\sqrt{14}z-26=0$$
So, the equations of the tangent planes are:
$$-\frac{1}{2}x-\frac{2}{3}y-2\sqrt{3}z-\frac{749}{36}=0$$
$$x-2y+2\sqrt{14}z-26=0$$
Key Concepts
Gradient VectorPartial DerivativesSurface Equation
Gradient Vector
The gradient vector is a critical concept in multivariable calculus, representing how a function changes in space. Whenever you have a surface described by an equation in terms of variables like \(x\), \(y\), and \(z\), the gradient vector helps us understand the direction and rate of steepest ascent on the surface.
To find the gradient vector of a function \(f(x, y, z)\), we calculate the partial derivatives with respect to each variable.
This yields a vector:
For the specific problem at hand, the equation is \(z^{2} - \frac{x^{2}}{16} - \frac{y^{2}}{9} - 1 = 0\). Differentiating with respect to all three variables, we obtain a gradient vector \(\langle -\frac{x}{8}, -\frac{2y}{9}, 2z \rangle\).
This tuple tells us how \(x\), \(y\), and \(z\) each contribute to changes on the surface.
To find the gradient vector of a function \(f(x, y, z)\), we calculate the partial derivatives with respect to each variable.
This yields a vector:
- \(abla f(x, y, z) = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle\)
For the specific problem at hand, the equation is \(z^{2} - \frac{x^{2}}{16} - \frac{y^{2}}{9} - 1 = 0\). Differentiating with respect to all three variables, we obtain a gradient vector \(\langle -\frac{x}{8}, -\frac{2y}{9}, 2z \rangle\).
This tuple tells us how \(x\), \(y\), and \(z\) each contribute to changes on the surface.
Partial Derivatives
Partial derivatives are the foundation for computing the gradient vector.
They reveal how a function changes as you change one of the variables while keeping the others constant.
In simpler terms, if you have a surface described by a function \(f(x, y, z)\), the partial derivative with respect to \(x\), \(\frac{\partial f}{\partial x}\), represents how \(f\) changes as \(x\) changes, holding \(y\) and \(z\) constant.
When calculating the tangent plane for the given problem, partial derivatives take center stage.
For the equation \(z^{2} - \frac{x^{2}}{16} - \frac{y^{2}}{9} - 1 = 0\), partial derivatives with respect to each variable were computed, resulting in:
This ultimately aids in finding the precise equation of the tangent plane at a particular point.
They reveal how a function changes as you change one of the variables while keeping the others constant.
In simpler terms, if you have a surface described by a function \(f(x, y, z)\), the partial derivative with respect to \(x\), \(\frac{\partial f}{\partial x}\), represents how \(f\) changes as \(x\) changes, holding \(y\) and \(z\) constant.
When calculating the tangent plane for the given problem, partial derivatives take center stage.
For the equation \(z^{2} - \frac{x^{2}}{16} - \frac{y^{2}}{9} - 1 = 0\), partial derivatives with respect to each variable were computed, resulting in:
- \(\frac{\partial f}{\partial x} = -\frac{x}{8}\)
- \(\frac{\partial f}{\partial y} = -\frac{2y}{9}\)
- \(\frac{\partial f}{\partial z} = 2z\)
This ultimately aids in finding the precise equation of the tangent plane at a particular point.
Surface Equation
The surface equation in question is \(z^{2} - \frac{x^{2}}{16} - \frac{y^{2}}{9} - 1 = 0\).
This equation describes a three-dimensional shape in space, a hyperbolic paraboloid, which can be visualized as a saddle-shaped surface.
The task is to find the tangent planes at two specific points on this surface.
To do this, we utilize the gradient vector derived from the surface's equation.
The equation of a tangent plane can be formulated using the gradient vector evaluated at the given points and the point-slope form.
For example, if the gradient at \( (4, 3, -\sqrt{3}) \) is \([-\frac{1}{2}, -\frac{2}{3}, -2\sqrt{3}]\), the tangent plane equation is derived as:
The process demonstrates the interplay of the surface equation's algebraic form with geometric interpretations.
This equation describes a three-dimensional shape in space, a hyperbolic paraboloid, which can be visualized as a saddle-shaped surface.
The task is to find the tangent planes at two specific points on this surface.
To do this, we utilize the gradient vector derived from the surface's equation.
The equation of a tangent plane can be formulated using the gradient vector evaluated at the given points and the point-slope form.
For example, if the gradient at \( (4, 3, -\sqrt{3}) \) is \([-\frac{1}{2}, -\frac{2}{3}, -2\sqrt{3}]\), the tangent plane equation is derived as:
- \(-\frac{1}{2}(x-4) - \frac{2}{3}(y-3) - 2\sqrt{3}(z+\sqrt{3}) = 0\)
The process demonstrates the interplay of the surface equation's algebraic form with geometric interpretations.
Other exercises in this chapter
Problem 15
Find all critical points of the following functions. $$f(x, y)=x^{4}-2 x^{2}+y^{2}-4 y+5$$
View solution Problem 15
Evaluate the following limits. $$\lim _{(x, y) \rightarrow(0, \pi)} \frac{\cos x y+\sin x y}{2 y}$$
View solution Problem 15
Computing gradients Compute the gradient of the following functions and evaluate it at the given point \(P\). $$F(x, y)=e^{-x^{2}-2 y^{2}} ; P(-1,2)$$
View solution Problem 15
Find the equation of the plane that is parallel to the vectors \langle 1,0,1\rangle and \(\langle 0,2,1\rangle,\) passing through the point (1,2,3)
View solution