A cable hangs in a curve that is symmetrical with respect to the y-axis.

Question

The tension, T (in Newtons) of the cable at a point x metres from the y-axis is given by 𝑇 = 200 (𝑒^x/10 + 𝑒^−𝑥/10). Find the maximum and minimum tension if the cable extends from 𝑥 = 10 to 𝑥 = −10 metres.

Answer 1

Given: 𝑇 = 200 (𝑒^𝑥/10 + 𝑒^−𝑥/10)

To find the minimum and maximum of the function T(x) in a given interval [-10,10] for x, first find the derivative of T and set it equal to zero.

T'(x) = 200(𝑒^𝑥/10 - 𝑒^−𝑥/10) = 0

Factor out 1/10

20(𝑒^𝑥- 𝑒^−𝑥) = 0

𝑒^𝑥- 𝑒^−𝑥 = 0

factor out 𝑒^-𝑥:

𝑒^-𝑥(𝑒^2𝑥-1) = 0

Set each factor equal to zero:

𝑒^-𝑥 = 0 | 𝑒^2𝑥-1=0

𝑒^-𝑥 = 0 has no real solution so:

𝑒^2𝑥=1

2x = ln (1)

2x = 0

x = 0

plugin to the function T:

T = 200(e⁰/10 + e⁰/10)

T = 200(1/10 + 1/10)=200(1/5)

T = 40 N (minimum ).

T(x) is symmetric at y-axis. If x→∞ or x →-∞, then T(x)→∞. Therefore the maximum value of T(x) is at x=±10

T= 200(e¹⁰/10 + e^-10/10)

T≈ 440,529.32 N (maximum)

Answer 2

Solution:

Step 1. Find the critical points of T on (-10, 10):

T(x) = 200 [e^(x/10) + e^(-x/10)]

T'(x) = 20 [e^(x/10) - e^(-x/10)]

Solve T'(x) = 0:

20 [e^(x/10) - e^(-x/10)] = 0

e^(x/10) = e^(-x/10)

e^(2x/10) = 1

x = 0

So the critical point is x=0.

Step 2: Compare T(0), T(10), and T(-10) to find the maximum and minimum values of T on [-10, 10]:

Since T is continuous on [-10, 10], it has maximum and minimum values on [-10, 10] and they are reached at either a critical point or an endpoint of the interval. Now

T(0) = 200 [e^(0/10 + e^(-0/10)] = 400

T(10) = T(-10) = 200 [e^(10/10 + e^(-10/10)] = 400 [e + e^(-1)] ≈ 617

But T(0) < T(10) = T(-10), so T has maximum value T(10) ≈ 617 and minimum value T(0)= 400.

Answer: The maximum tension is 617 newtons and the minimum tension is 400 newtons.