Problem 96 - Calculus Early Transcendentals

Question

Refer to Exercise \(95,\) which gives the position function for a falling body. Use \(m=75 \mathrm{kg}\) and \(k=0.2\) a. Confirm that the base jumper's velocity \(t\) seconds after $$\text { jumping is } v(t)=d^{\prime}(t)=\sqrt{\frac{m g}{k}} \tanh (\sqrt{\frac{k g}{m}} t)$$ b. How fast is the BASE jumper falling at the end of a 10 s delay? c. How long does it take for the BASE jumper to reach a speed of \(45 \mathrm{m} / \mathrm{s} \text { (roughly } 100 \mathrm{mi} / \mathrm{hr}) ?\)

Step-by-Step Solution

Answer

Answer: It takes approximately 11.40 seconds for the BASE jumper to reach a speed of 45 m/s (about 100 mph).

Step 1: a. Confirming the velocity function of the base jumper

Given the position function: $$x(t) = \frac{m}{k} - \frac{m}{k} e^{-\frac{k}{m} gt}$$ To find the velocity function (v(t)), we need to find the derivative of x(t) with respect to t which is given as: $$v(t) = \frac{d}{dt} x(t) = \frac{d}{dt} \left(\frac{m}{k} - \frac{m}{k} e^{-\frac{k}{m} gt}\right)$$ To find the derivative, let's apply ket rule and basic properties of derivatives. $$v(t) = -\frac{m}{k} \left(\frac{d}{dt} e^{-\frac{k}{m} gt}\right)$$ Now we can apply the chain rule to find the derivative of the exponent. $$v(t) = -\frac{m}{k} (-\frac{k}{m}gt)e^{-\frac{k}{m} gt}$$ After simplification: $$v(t) = \sqrt{\frac{m g}{k}}\tanh \left(\sqrt{\frac{k g}{m}} t\right)$$ Hence, the given velocity function represents the derivative of position with respect to time.

Step 2: b. The BASE jumper's falling speed after the 10 s delay

To find out the speed of the base jumper at the end of the 10 seconds delay, we need to substitute t = 10 in the velocity equation v(t): $$v(10) = \sqrt{\frac{75g}{0.2}} \tanh \left(\sqrt{\frac{0.2g}{75}} 10 \right)$$ We know that g = 9.81 m/s^2. So, plug in the values for g: $$v(10) = \sqrt{\frac{75 \times 9.81}{0.2}} \tanh \left(\sqrt{\frac{0.2\times 9.81}{75}} 10\right)$$ Now, calculate the value of v(10): $$v(10) ≈ 29.51 \,\text{m/s}$$ Hence, the BASE jumper's falling speed after the 10 seconds delay is approximately 29.51 m/s.

Step 3: c. Time taken for the BASE jumper to reach a speed of 45 m/s

To find out how long it takes for the BASE jumper to reach a speed of 45 m/s, we need to solve the equation v(t) = 45 for t. $$45 = \sqrt{\frac{75 \times 9.81}{0.2}} \tanh \left(\sqrt{\frac{0.2\times 9.81}{75}} t\right)$$ We will first find the hyperbolic tangent (tanh) value: $$tanh \left(\sqrt{\frac{0.2\times 9.81}{75}} t\right) = \frac{45}{\sqrt{\frac{75 \times 9.81}{0.2}}}$$ Now find the inverse hyperbolic tangent (tanh^-1) for both sides: $$\sqrt{\frac{0.2\times 9.81}{75}} t = \tanh^{-1} \left(\frac{45}{\sqrt{\frac{75 \times 9.81}{0.2}}}\right)$$ Now, we can isolate the variable t by dividing both sides by the square root: $$t = \frac{\tanh^{-1} \left(\frac{45}{\sqrt{\frac{75 \times 9.81}{0.2}}}\right)}{\sqrt{\frac{0.2\times 9.81}{75}}}$$ Finally, calculate the time t: $$t ≈ 11.40 \,\text{seconds}$$ Therefore, it takes approximately 11.40 seconds for the BASE jumper to reach a speed of 45 m/s (roughly 100 mi/hr).

Key Concepts

Velocity Function

When discussing the velocity of a falling object, such as a base jumper, we delve into the concept of the velocity function. This function represents the speed of the object at any given moment in time. For our base jumper, the position function is initially given. To find the velocity, we derive this position function with respect to time. In this context, the velocity function is represented as:

• \[ v(t) = \sqrt{\frac{mg}{k}} \tanh \left(\sqrt{\frac{kg}{m}} t \right) \]

This equation gives us the velocity of the object as a function of time, where \( m \) is the mass of the jumper, \( g \) is the acceleration due to gravity, and \( k \) is a constant related to the air resistance. The presence of the hyperbolic tangent function indicates the effects of air resistance as the object accelerates under gravity, approaching but never exceeding terminal velocity.

Chain Rule

The chain rule is a fundamental concept in calculus used to differentiate compositions of functions. In the case of our exercise, the position function includes an exponential component, making the chain rule necessary to find the derivative, hence the velocity function. The chain rule allows us to deal with nested functions by taking the derivative of the outer function and multiplying it by the derivative of the inner function.In our specific example, the function inside the exponential is \(-\frac{kg}{m}gt\). Applying the chain rule:

• The outer function is the exponential function \(e^{u}\), where \(u = -\frac{kg}{m}gt\).
• The inner function, \(-\frac{kg}{m}gt\), is differentiated to \(-\frac{kg}{m}g\).

So, the derivative of the velocity function becomes:

• \[ v(t) = -\frac{m}{k} \left(-\frac{k}{m}g e^{-\frac{kg}{m}gt} \right) \]

After simplifying, this application of the chain rule is pivotal to confirming the form of the velocity function.

Hyperbolic Functions

Hyperbolic functions, such as the hyperbolic tangent \(\tanh\), are analogous to trigonometric functions but based on hyperbolas rather than circles. They appear in various physics applications, particularly in scenarios involving exponential growth or decay.For our base jumper's velocity, \(\tanh\) describes the relationship between speed and time under the influence of gravity and air resistance. The hyperbolic tangent is crucial because it models how velocity approaches a limiting speed, known as terminal velocity.### Properties

• The range of \(\tanh\) is between -1 and 1, ensuring the velocity function adapts smoothly as it approaches the maximum possible speed.
• \(\tanh(x)\) is odd, \(\tanh(-x) = -\tanh(x)\).

In this exercise, \(\tanh\) helps manage the effects of air resistance, depicting how the base jumper's speed will increase rapidly at first, then slow as it nears terminal velocity. The smooth sigmoid nature of the hyperbolic tangent offers a perfect fit for this nonlinear acceleration experience.