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Herbie S.
PhD Student in Physics at The University of Texas at Austin
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Question:

Suppose you have just discovered the early retirement movement and you want to calculate how much you will need in investments in order to retire. Assume your monthly expenses are $3,500 and that you choose an annual withdrawal rate of 3% of your total investments to support your living expenses. How much total investment assets do you need to live off this amount forever? Herbie S. Answer: First, convert monthly expenses to annual expenses by taking ($3,500/month)(12 months/yr) = $42,000/year. The total amount you need invested is related to total annual expenses and the interest rate by (Total investments)(Interest rate) = Annual expenses. We can reorganize this equation to give us the total investments needed: Total investments = (Annual expenses)/(Interest rate) = ($42,000)/(0.03) Total investments = $1,400,000 Thus, you will need$1,400,000 to safely retire with monthly expenses of $3,500. Calculus TutorMe Question: Consider the classic problem of a ladder leaning on a wall. The ladder is 20 feet long and the bottom of the ladder is 12 feet from the wall. You begin pushing the bottom of the ladder toward the wall at a rate of 3 feet per second. How quickly does the top of the ladder move up the wall? Herbie S. Answer: This problem is a type of related rates problem, meaning that the rate that the top of the ladder moves is related to the rate at which the bottom of the ladder moves. In this case, the ground, the wall, and the ladder make up three sides of a triangle, so the distances are related by the Pythagorean Theorem,$( x^2 + y^2 = h^2. $) Furthermore, the relationship between the rates at which the top and bottom of the ladder move along the wall is given by differentiating the Pythagorean Theorem with respect to $$t$$. Taking the derivative gives us$( 2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 2h \frac{dh}{dt}. $) Now, let's assign the numbers in the problem statement to variables. We'll label distance along the ground by $$x$$, distance along the wall by $$y$$, and distances along the ladder by $$h$$. Using these conventions, we can quickly pull from the problem statement that $$x = 12,$$ $$h = 20,$$ and $$\frac{dx}{dt} = -3$$. The negative sign arises because we are pushing the bottom of the ladder toward the wall, reducing the distance between the ladder and the wall. Since the ladder isn't getting any longer or shorter, we also have that $$\frac{dh}{dt} = 0.$$ Next, we can plug these values into the Pythagorean Theorem and its derivative to find $$y$$ and $$\frac{dy}{dt}$$. Plugging known values for $$x$$ and $$h$$ into the Pythagorean Theorem gives us$( 12^2 + y^2 = 20^2 $)$( y = \sqrt{20^2-12^2} = 16.$) Finally, we plug all of our known values into the derivative of the Pythagorean Theorem to find the solution for $$\frac{dy}{dt}$$:$( 2(12)(-3) + 2(16)\frac{dy}{dt} = 2(20)(0). $)$( \boxed{\frac{dy}{dt} = \frac{9}{4}.} $) So the top of the ladder moves up the wall at $$\frac{9}{4} \frac{ft}{s}.$$ Physics TutorMe Question: Consider a block of mass $$m$$ held at rest at a height of $$h$$ meters. How long does it take the block to fall to the ground once it is released? Assume we're on Earth, so $$g = -9.8 \frac{m}{s^2}$$. Herbie S. Answer: Because this problem only involves a constant acceleration due to gravity, we can use the kinematic equations to solve it. Recall that the kinematic equation is$(x_f = \frac{1}{2} a t^2 + v_0 t+ x_0. $) In this case, we can quickly plug in a few known quantities from the problem statement. The acceleration in the kinematic equation is just equal to the acceleration due to gravity, so $$a=g,$$ and the initial velocity of the block is equal to 0 because it is being held in place and then released, so $$v_0 = 0$$. The initial position is equal to the height at which the block is held, meaning $$x_0 = h$$, and the final position should simply be 0, meaning we have reached the ground, so $$x_f = 0$$. Plugging these values into the kinematic equation gives us a simpler equation to work with:$( 0 = \frac{1}{2} g t^2 + h. $) Finally, to find how long it takes the block to reach the ground, we solve for $$t$$ algebraically. Our final result is$( \boxed{t = \sqrt{-\frac{2h}{g}}} \$). As an added bonus, notice that this result doesn't depend on the mass of the block, which we called $$m$$. This shows that if we were to carry out an experiment where we drop a feather and a bowling ball at the same time in a vacuum, where there's no drag force because there's no air to push on the objects, we can expect them to hit the ground at the same time!

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