[SOLVED] MATH2263 Project #1- Comparing Surfaces

1 Part 1: Traces

The cone and paraboloid are two mathematical surfaces with distinct shapes that, at first glance, may appear to be similar. A cursory examination of both Figure 1 and Figure 2 reveals that both shapes exhibit an almost circular-like distribution. However, upon delving deeper into this project, I will demonstrate how significantly different these shapes are.

Upon observing the contour maps of the cone and paraboloid, a notable disparity in their growth patterns is immediately discernible. Specifically, the paraboloid exhibits a rate of growth that could be described as approaching an exponential function, evidenced by the discrepancy in growth between z = 4 and z = 3 when compared to the difference in growth between z = 2 and z = 3.

In contrast, the growth of the cone appears to be more uniform across its di- mensions. This observation could be attributed to the paraboloid’s underlying quadratic equation, which imparts a degree of curvature and accentuates the

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rate of growth in the z direction. Such distinctions are critical to understand- ing the properties and applications of these shapes in various fields, including physics, engineering, and mathematics.

2 Part 2: Directional Derivatives

I began by calculating the directional derivatives and magnitude of both shapes:

2.1 Step 1

Find the Gradient Vector and Magnitude for z = x2 + y2: RecalltheGradientVectorFormula: ∇f=􏰀fx,fy,fz􏰁=∂fi+∂fj+∂fk

∂x ∂y ∂z

∂f =2x, ∂f =2y, ∂f =−1−→∇f =􏰀2x,2y,−1􏰁 ∂x ∂y ∂z

Plugging in the point (1, 0, 1), we have: ∇f(1,0,1) = 􏰀2,0,−1􏰁

Gradient Vector: 􏰀2,0,−1􏰁 􏰃􏰀􏰁􏰃􏰄 √

􏰃 2,0,−1 􏰃= 22 +02 +(−1)2 = 5 √

Magnitude: = 5

2.2 Step 2

Find the Gradient Vector and Magnitude for z = 􏰄x2 + y2: RecalltheGradientVectorFormula: ∇f=􏰀fx,fy,fz􏰁=∂fi+∂fj+∂fk

∂x ∂y ∂z

∂f x ∂f y ∂f 􏰀 x y 􏰁 ∂x = 􏰄x2 +y2, ∂y = 􏰄x2 +y2, ∂z =−1−→∇f = 􏰄x2 +y2,􏰄x2 +y2,−1

Plugging in the point (1, 0, 1), we have:

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2.3 Final Answer

∇f(1,0,1) = 􏰀1,0,−1􏰁 Gradient Vector: 􏰀1,0,−1􏰁

􏰃􏰀􏰁􏰃􏰄 √ 􏰃 1,0,−1 􏰃= 12 +02 +(−1)2 = 2

√

Magnitude: = 2

Upon analyzing the rate of change of the surface areas of a paraboloid and a cone, it is evident that the paraboloid has the highest rate of increase, with a

√√

value of 5, while the cone has a rate of increase of 2. The computation of the greatest increase highlights the substantial difference between the two shapes, revealing that the paraboloid increases at a much faster rate than the cone.

Furthermore, observing the contour maps for both shapes provides a graph- ical representation of their rate of change. From the contour map, it is clear that the paraboloid has a much larger increase as compared to the cone, while the latter seems to increase at a relatively constant rate.

3 Part 3: Tangent Places

3.1 Step 1

Find the Tangent Planes on P (0, 0, 0) for z = x2 + y2 ∂f =2x, ∂f =2y,

∂x ∂y Plugging in x = 0 and y = 0, we get:

z − 0 = 2x(0, 0)(x − 0) + 2y(0, 0)(y − 0) 3

so the equation of the tangent plane is:

z=0

3.2 Step 2

∂x=􏰄x2+y2, ∂y=􏰄x2+y2, Plugging in x = 0 and y = 0, we get:

xy
z − 0 = 􏰄x2 + y2 (0, 0)(x − 0) + 􏰄x2 + y2 (0, 0)(y − 0)

so the equation of the tangent plane is:

z = 0.

3.3 Step 3

Plotting the surface and the tangent

3DimageofbothZ=􏰄x2+y2 andZ=x2+y2

To provide a comprehensive understanding of the cone and paraboloid shapes, I decided to utilize the powerful visualization tool, GeoGebra 3D. This tool al- lows for an interactive 3D representation of the shapes, which provides a better insight into their structure and properties.

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Find the Tangent Planes on P (0, 0, 0) for Z = 􏰄x2 + y2 ∂f x ∂f y

3.4 Final Answer

Therefore, in our effort to derive the equation of the tangent plane to the surface z = f(x,y) at the point P(0,0,0), we have obtained z = 0 as the equation of the tangent plane. This result indicates that the tangent line at the origin lies on the x-y plane. This finding can also be clearly observed in the accompanying graphs (figure 1 & 2), where the point P (0, 0, 0) is seen to be untouched by the surface.

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