Math Problem Statement

A right circular cone of height h and base radius r has total surface area S consisting of its base area plus its side area, leading to the formula:

S= πœ‹ r^2 + πœ‹r * sqrt(r^2+h^2)

Suppose you start out with a cone of height 8 cm and base radius 6 cm, and you want to change the dimensions in such a way that the total surface area remains the same. Suppose you increase the height by 32/100. In this problem, use tangent line approximation to estimate the new value of r so that the new cone has the same total surface area.

The estimated value of r =

Solution

To solve this problem, we'll use the concept of tangent line approximation, which is essentially linear approximation. Here’s the step-by-step approach:

Step 1: Define the Surface Area Function

The formula for the surface area SS of a cone with height hh and base radius rr is given by: S=Ο€r2+Ο€rr2+h2S = \pi r^2 + \pi r \sqrt{r^2 + h^2}

Step 2: Set Up the Original Values and Changes

Given:

  • Initial height h=8h = 8 cm
  • Initial radius r=6r = 6 cm
  • Increase in height Ξ”h=0.32Γ—h=0.32Γ—8=2.56\Delta h = 0.32 \times h = 0.32 \times 8 = 2.56 cm

Thus, the new height hβ€²=h+Ξ”h=8+2.56=10.56h' = h + \Delta h = 8 + 2.56 = 10.56 cm.

Step 3: Find the Derivative of SS with Respect to rr

Since we're using tangent line approximation, we need the partial derivatives of SS with respect to rr and hh.

The derivative of SS with respect to rr, denoted βˆ‚Sβˆ‚r\frac{\partial S}{\partial r}, is: βˆ‚Sβˆ‚r=2Ο€r+Ο€r2+h2+Ο€r2r2+h2\frac{\partial S}{\partial r} = 2\pi r + \pi \sqrt{r^2 + h^2} + \frac{\pi r^2}{\sqrt{r^2 + h^2}}

Step 4: Find the Derivative of SS with Respect to hh

The derivative of SS with respect to hh, denoted βˆ‚Sβˆ‚h\frac{\partial S}{\partial h}, is: βˆ‚Sβˆ‚h=Ο€rhr2+h2\frac{\partial S}{\partial h} = \frac{\pi r h}{\sqrt{r^2 + h^2}}

Step 5: Apply Linear Approximation

For the surface area SS to remain the same, the total differential dSdS must be zero: dS=βˆ‚Sβˆ‚r dr+βˆ‚Sβˆ‚h dh=0dS = \frac{\partial S}{\partial r} \, dr + \frac{\partial S}{\partial h} \, dh = 0

Thus, βˆ‚Sβˆ‚r dr=βˆ’βˆ‚Sβˆ‚h dh\frac{\partial S}{\partial r} \, dr = - \frac{\partial S}{\partial h} \, dh

Solving for drdr (the change in rr): dr=βˆ’βˆ‚Sβˆ‚h dhβˆ‚Sβˆ‚rdr = -\frac{\frac{\partial S}{\partial h} \, dh}{\frac{\partial S}{\partial r}}

Step 6: Substitute Values

Substitute r=6r = 6, h=8h = 8, and dh=2.56dh = 2.56 into the derivatives and calculate drdr. Then, find the new radius rβ€²=r+drr' = r + dr.

I'll perform the calculations for drdr and find the approximate value of the new radius.The estimated new value of the radius rr to keep the surface area constant after increasing the height is approximately rβ‰ˆ5.52r \approx 5.52 cm.

Would you like further details or have any questions?

Here are some additional related questions for practice:

  1. How would the result change if the height increased by 50% instead of 32%?
  2. Can you derive a general formula for the radius change drdr in terms of dhdh, rr, and hh?
  3. What is the volume of the cone before and after the height increase, using the new radius?
  4. How does the approximation accuracy vary if we increase or decrease both rr and hh proportionally?
  5. What happens to the surface area if only the height changes without adjusting the radius?

Tip: Tangent line approximations work best for small changes in variables. For larger changes, other methods may provide more accurate results.

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Math Problem Analysis

Mathematical Concepts

Calculus
Linear Approximation
Differential Calculus
Geometry of Cones

Formulas

Surface Area of Cone S = Ο€ r^2 + Ο€ r * sqrt(r^2 + h^2)
Linear Approximation dS = (βˆ‚S/βˆ‚r) dr + (βˆ‚S/βˆ‚h) dh

Theorems

Differential Approximation (Tangent Line Approximation)

Suitable Grade Level

Grades 11-12