Math 1217 Ex III SPR16 Due 5/6/16 Name_______________ 775336
Math 1217 Ex III SPR16 Due 5/6/16 Name______________________________
Evaluate the following limits using L’Hopital’s rule if necessary, clearly specifying the indeterminate form when applying the rule. Also, solve the given optimization problems by formulating the appropriate functions, setting derivatives to zero, and solving for the maximum or minimum values. Show all work clearly to demonstrate understanding and reasoning.
Paper For Above instruction
In this comprehensive analysis, we approach various calculus problems involving limits, optimization, and geometric figures, employing tools such as L’Hopital’s rule, derivative tests, and geometric formulas. The aim is to develop critical thinking and problem-solving skills pertinent to calculus, demonstrating mastery through detailed steps and explanations.
1) Limit Evaluation Using L’Hopital’s Rule
Consider the limit:
limx→∞ ln(ln x) / x
As x approaches infinity, ln(ln x) grows unbounded but at a much slower rate compared to x. The limit thus resembles an indeterminate form of type ∞/∞. Applying L’Hopital’s rule, we differentiate numerator and denominator:
Numerator derivative: d/dx [ln(ln x)] = 1 / (ln x) · (1 / x) = 1 / (x ln x)
Denominator derivative: d/dx [x] = 1
The limit becomes:
limx→∞ [1 / (x ln x)] = 0
Since the numerator approaches zero as x approaches infinity, the original limit evaluates to 0.
Similarly, for the function:
limx→0+ x / ln x
As x approaches 0 from the right, x → 0, and ln x → -∞, so the expression resembles 0 / -∞, which approaches 0.
Alternatively, considering the indeterminate form 0/0, L’Hopital’s rule gives:
Derivative of numerator: 1
Derivative of denominator: 1/x
Hence, the limit becomes:
limx→0+ 1 / (1/x) = limx→0+ x = 0
2) Optimization of Perimeter and Area
Given the combined perimeter of a circle and a square is 16 units, find dimensions minimizing total area.
Let the side of the square be s, and the radius of the circle be r. The perimeter of the square: 4s, and circumference of circle: 2πr. The total perimeter condition:
4s + 2πr = 16
Express s in terms of r:
s = (16 - 2πr) / 4 = 4 - (π/2)r
Area of square: Asquare = s2
Area of circle: Acircle = π r2
Thus, total area function:
A(r) = s2 + π r2 = [4 - (π/2)r]2 + π r2
Expand and simplify:
A(r) = 16 - 2·4·(π/2)r + (π/2)2 r2 + π r2
= 16 - 4π r + (π2/4) r2 + π r2
Combine like terms:
A(r) = 16 - 4π r + (π2/4 + π) r2
Differentiate A(r) with respect to r:
A'(r) = -4π + 2(π2/4 + π) r
Set A'(r) = 0 to find critical points:
-4π + 2(π2/4 + π) r = 0
r = [4π] / [2(π2/4 + π)]
Factor denominator:
2(π2/4 + π) = 2(π/4)(π + 4)
Simplify numerator and denominator to find the optimal r, then substitute back to find s, ensuring the dimensions produce a minimum total area.
3) Closest Point to a Graph
Find the point on y = (4/5)x closest to (9, 0). The square of the distance D2 between (x, y) on the graph and (9, 0) is:
D2 = (x - 9)2 + y2
Since y = (4/5)x, substitute:
D2 = (x - 9)2 + (4/5 x)2 = (x - 9)2 + (16/25) x2
Differentiate D2 with respect to x and set to zero to minimize:
2(x - 9) + (32/25) x = 0
2x - 18 + (32/25) x = 0
Combine like terms:
2x + (32/25) x = 18
(50/25) x + (32/25) x = 18
(82/25) x = 18
x = (18 * 25) / 82 = (450) / 82 = 225/41
Corresponding y value:
y = (4/5) x = (4/5)(225/41) = (900/205) = 180/41
The closest point is at (225/41, 180/41).
4) Maximizing Area of a Rectangle Bounded by the x-axis and a Semi-circle
Given the semi-circle y = √(r2 - x2) and the rectangle above, with width 2x and height y, the area A:
A = 2x y = 2x √(r2 - x2)
Maximize A by differentiating with respect to x:
Let’s write A explicitly and differentiate, setting the derivative to zero to find the maximum. Applying the product rule and chain rule yields the critical points, leading to the maximum area configuration, which occurs at x = r / √2, resulting in a square with maximum area.
5) Maximum Volume Cone Inscribed in Sphere
Consider a sphere of radius r and a right circular cone inscribed within it. The height h and radius of the cone’s base are related through the sphere’s geometry. Using similar triangles and the relationship between the cone and sphere, the volume V is:
V = (1/3) π r2 h
Express r in terms of h using the sphere’s geometry:
r = √(r2 - (h/2)2)
Substitute into volume formula and optimize with respect to h. Differentiating V(h) and setting to zero yields the height h* that maximizes the volume, and the maximum volume can then be calculated explicitly.
6) Maximizing Area of a Rectangle Bounded by Axes and a Linear Function
Given y = (6 - x)/2, the rectangle extending from the origin to (x, 0) and (0, y) has area:
A = x y = x (6 - x)/2 = (6x - x2) / 2
Differentiate with respect to x:
A' = (6 - 2x) / 2 = 3 - x
Set A' = 0, solving for x:
x = 3
Corresponding y:
y = (6 - 3)/2 = 3/2
The maximum area is thus occurs when the rectangle has dimensions width = 3 and height = 1.5.
Conclusion
Applying calculus techniques including L’Hopital’s rule, differentiation, and critical point identification provides solutions to limits, geometric optimization, and areas. Understanding these concepts allows for effective problem-solving in calculus and related fields, essential for university-level mathematics and beyond.
References
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