Problem 26
Question
A bee sat at the point \((1,2,1)\) on the ellipsoid \(x^{2}+y^{2}+2 z^{2}=6\) (distances in feet). At \(t=0\), it took off along the normal line at a speed of 4 feet per second. Where and when did it hit the plane \(2 x+3 y+z=49 ?\)
Step-by-Step Solution
Verified Answer
The bee hits the plane at \(\left(\frac{343}{63}, \frac{704}{63}, \frac{681}{63}\right)\) at \(\frac{40}{63}\) seconds.
1Step 1: Identify the normal vector
The normal vector to the ellipsoid at point \(1,2,1\) is the gradient of the ellipsoid equation at that point. The gradient is found by taking the partial derivatives of the ellipsoid equation, \(x^2+y^2+2z^2=6\), with respect to \(x\), \(y\), and \(z\). So, \([2x, 2y, 4z]\). Evaluating at \(x=1\), \(y=2\), \(z=1\), the normal vector is \([2(1), 2(2), 4(1)] = [2, 4, 4]\)."
2Step 2: Write the parametric equations
The parametric equations for the bee's path (normal line) using the point \(1, 2, 1\) and the normal vector \([2, 4, 4]\), with \(t\) representing time and the speed of 4 feet per second are: \(x=1 + 8t\), \(y=2 + 16t\), \(z=1 + 16t\). The coefficients \(8\) and \(16\) come from multiplying the components of the direction vector \([2, 4, 4]\) by the speed 4."
3Step 3: Find when the path intersects the plane
Substitute the parametric equations into the plane equation \(2x + 3y + z = 49\). \[ 2(1 + 8t) + 3(2 + 16t) + (1 + 16t) = 49 \]Solve this equation for \(t\). \[ 2 + 16t + 6 + 48t + 1 + 16t = 49 \]\[ 63t + 9 = 49 \]\[ 63t = 40 \]\[ t = \frac{40}{63} \].This value of \(t\) represents the time in seconds at which the bee hits the plane.
4Step 4: Determine the hit point on the plane
Plug \(t = \frac{40}{63}\) back into the parametric equations to find the coordinates of the point of impact:For \(x\): \(x = 1 + 8\left(\frac{40}{63}\right) = \frac{343}{63}\).For \(y\): \(y = 2 + 16\left(\frac{40}{63}\right) = \frac{704}{63}\).For \(z\): \(z = 1 + 16\left(\frac{40}{63}\right) = \frac{681}{63}\)."
Key Concepts
Normal VectorParametric EquationsGradientPlane Intersection
Normal Vector
A normal vector is a vector that is perpendicular to a surface at a given point. In the context of ellipsoids, the normal vector is critically important as it defines the direction of the steepest ascent or descent on the surface.
When dealing with a bee sitting on an ellipsoid, like the one described by the equation \(x^2+y^2+2z^2=6\), we can pinpoint the direction in which it initially flies when taking off. This direction is along the normal vector.
To find the normal vector at the point \((1,2,1)\), we calculate the gradient of the ellipsoid's equation. The gradient is given by the partial derivatives of the equation and this results in the vector \([2x, 2y, 4z]\).
Evaluating these at the point \((1,2,1)\), we find the normal vector as \([2(1), 2(2), 4(1)] = [2, 4, 4]\). This vector signifies the bee's direction of travel as it takes off from the ellipsoid.
When dealing with a bee sitting on an ellipsoid, like the one described by the equation \(x^2+y^2+2z^2=6\), we can pinpoint the direction in which it initially flies when taking off. This direction is along the normal vector.
To find the normal vector at the point \((1,2,1)\), we calculate the gradient of the ellipsoid's equation. The gradient is given by the partial derivatives of the equation and this results in the vector \([2x, 2y, 4z]\).
Evaluating these at the point \((1,2,1)\), we find the normal vector as \([2(1), 2(2), 4(1)] = [2, 4, 4]\). This vector signifies the bee's direction of travel as it takes off from the ellipsoid.
Parametric Equations
Parametric equations are a way of expressing geometric shapes using parameters. They help in describing paths or lines in space using variables that depend on one or more parameters.
In this exercise, since the bee takes off along the normal line, we represent this path through parametric equations. This involves using the initial point \((1,2,1)\) and the previously determined normal vector \([2, 4, 4]\).
Given a speed of 4 feet per second, the parametric equations become \(x=1 + 8t\), \(y=2 + 16t\), \(z=1 + 16t\). We get these by multiplying the components of the direction vector \([2, 4, 4]\) by this speed. The variable \(t\) denotes time in seconds, showing how locational coordinates change as time progresses.
In this exercise, since the bee takes off along the normal line, we represent this path through parametric equations. This involves using the initial point \((1,2,1)\) and the previously determined normal vector \([2, 4, 4]\).
Given a speed of 4 feet per second, the parametric equations become \(x=1 + 8t\), \(y=2 + 16t\), \(z=1 + 16t\). We get these by multiplying the components of the direction vector \([2, 4, 4]\) by this speed. The variable \(t\) denotes time in seconds, showing how locational coordinates change as time progresses.
Gradient
The gradient is a crucial concept in calculus and vector mathematics. It acts as a multi-variable measure of a function's rate of change and direction.
For three variables \(x\), \(y\), and \(z\), the gradient consists of the partial derivatives \([ \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} ]\). This forms a vector that tells us the direction of maximum increase of a function.
In our exercise context, the gradient \([2x, 2y, 4z]\), derived from the ellipsoid function \(x^2+y^2+2z^2=6\), determines the direction of the normal vector. Calculating this at point \((1,2,1)\) reveals how the bee sets off in the direction of greatest ascent on the ellipsoid.
For three variables \(x\), \(y\), and \(z\), the gradient consists of the partial derivatives \([ \frac{\partial}{\partial x}, \frac{\partial}{\partial y}, \frac{\partial}{\partial z} ]\). This forms a vector that tells us the direction of maximum increase of a function.
In our exercise context, the gradient \([2x, 2y, 4z]\), derived from the ellipsoid function \(x^2+y^2+2z^2=6\), determines the direction of the normal vector. Calculating this at point \((1,2,1)\) reveals how the bee sets off in the direction of greatest ascent on the ellipsoid.
Plane Intersection
The intersection of a path with a plane is where a moving object meets a stationary impenetrable surface. In our bee's adventure, it flies along a parametric path until it intersects with a plane given by \(2x + 3y + z = 49\).
To find this point of intersection, we substitute our path's parametric equations into the plane equation. Replacing \(x\), \(y\), and \(z\) with \(1 + 8t\), \(2 + 16t\), and \(1 + 16t\) respectively, we solve for \(t\).
The calculation \[2(1 + 8t) + 3(2 + 16t) + (1 + 16t) = 49\] simplifies to \[63t + 9 = 49\]. Solving gives \(t = \frac{40}{63}\), the time in seconds at which the bee hits the plane. By plugging \(t = \frac{40}{63}\) back into the parametric equations, we determine exact coordinates of intersection: \(x = \frac{343}{63}\), \(y = \frac{704}{63}\), \(z = \frac{681}{63}\).
To find this point of intersection, we substitute our path's parametric equations into the plane equation. Replacing \(x\), \(y\), and \(z\) with \(1 + 8t\), \(2 + 16t\), and \(1 + 16t\) respectively, we solve for \(t\).
The calculation \[2(1 + 8t) + 3(2 + 16t) + (1 + 16t) = 49\] simplifies to \[63t + 9 = 49\]. Solving gives \(t = \frac{40}{63}\), the time in seconds at which the bee hits the plane. By plugging \(t = \frac{40}{63}\) back into the parametric equations, we determine exact coordinates of intersection: \(x = \frac{343}{63}\), \(y = \frac{704}{63}\), \(z = \frac{681}{63}\).
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