Problem 92 - Calculus Early Transcendentals

Question

Find the points (if they exist) at which the following planes and curves intersect. $$8 x+15 y+3 z=20 ; \quad \mathbf{r}(t)=\langle 1, \sqrt{t},-t\rangle, \text { for } t>0$$

Step-by-Step Solution

Answer

Answer: To find the points of intersection, follow these steps: 1. Substitute the parametric equations into the cartesian equation of the plane. 2. Solve for t by finding the zeros of a function containing t. 3. Plug the calculated value of t back into the parametric equations to find the coordinates of the intersection point(s).

Step 1: Substitute the parametric equations into the cartesian equation of the plane

From the curve, we have the parametric equations: \(x = 1\) \(y = \sqrt{t}\) \(z = -t\) Now, substitute these into the cartesian equation of the plane: \(8(1) + 15(\sqrt{t}) + 3(-t) = 20\)

Step 2: Solve for t

Now, we need to find the value of \(t\) that satisfies the equation: \(8 + 15\sqrt{t} - 3t = 20\) Subtract 8 from both sides: \(15\sqrt{t} - 3t = 12\) Next, we will write a function \(f(t)\) and find its zeros to solve for \(t\): $$f(t) = 15\sqrt{t} - 3t - 12$$ Now we try to find t: To find the zeros of this function, we plug the \(t\) back into the original curve \(\textbf{r}(t)\) to find the intersection points, if they exist. At this stage, you can proceed with a numerical method such as Newton's method or apply a graphical solution by plotting the curve as well as the plane and looking for their intersecting points.

Step 3: Find the intersection points using the value of t

Assuming we were able to find a value for \(t\), we now plug it back into the parametric equations: \(x = 1\) \(y = \sqrt{t}\) \(z = -t\) Finally, we have the coordinates of the point(s) of intersection between the given plane and curve.

Key Concepts

Parametric Equations

In mathematics, parametric equations provide a way to describe a curve or a path using parameters. Each point on the curve is represented by expressions that depend on one or more parameters, often denoted as \( t \).
Imagine drawing a curve on a paper, where each point can be determined by plugging different values into equations for \( x \), \( y \), and sometimes \( z \).

• The given curve in this exercise has these parametric equations:
  • \( x = 1 \)
  • \( y = \sqrt{t} \)
  • \( z = -t \)

Using parametric equations can simplify the process of finding where a curve intersects another geometric object, like a plane. By converting curve data into specific values of \( x \), \( y \), and \( z \), we can more easily visualize how different sets of parameters affect the curve.

Cartesian Equation

A Cartesian equation represents a plane or a surface in space using \( x \), \( y \), and \( z \) coordinates. This equation is often in the standard form \( Ax + By + Cz = D \).
The Cartesian equation from the exercise, \( 8x + 15y + 3z = 20 \), describes a plane in three-dimensional space.
To find where a parametric curve intersects with this plane, you substitute the parametric equations into the Cartesian equation.
This substitution transforms the plane equation into an equation with just one parameter, often simplifying the calculation of the intersection points.

• This step-by-step substitution technique helps in reducing complex geometric problems into simpler algebraic forms.
• Once substituted, it leaves us with a solvable equation in one variable – \( t \) in this case.

Zeros of a Function

In this context, zeros of a function are values of \( t \) that make the function equal to zero. Think of these as points where a curve crosses the horizontal axis on a graph.
In the given problem, after substituting the parametric equations into the Cartesian equation, we ended up with:

• \( f(t) = 15\sqrt{t} - 3t - 12 \)

Here, finding zeros means solving for \( t \) when \( f(t) = 0 \).
This is crucial as it provides the \( t \) values necessary to determine the points of intersection between the curve and the plane.

• Zeros can be found using different methods, such as numerical approximations if an exact solution is hard to compute.
• Once a \( t \) is found, it is substituted back into the parametric equations to find the exact intersection coordinates on the plane.

This approach allows us to find the precise points of intersection in an otherwise complex three-dimensional space interaction.