Question

3.Two posts, one 8ft high and the other 12ft high, stand 15ft apart. They are to be stayed by wires attached to a single stake at ground level, the wires running to the tops of the posts. Where the stake should be placed, to use the least amount of wire?
1.At a given instant the legs of a right triangle are 8 in., and 6 in., respectively. The first leg decreases at 1 in/min and the second increases at 2 in/min. At what rate is the area increasing after 2 min.?
2.An arc light is 15 feet above a sidewalk. A man 6 feet tall walks away from the point under the light at the rate of 5ft/sec. How fast is his shadow lengthening when he is 20 feet away from the point under the light?
3.A man starts walking eastward at 5ft/sec from a point A. Ten minutes later a second man starts walking west at the rate of 5ft/sec from a point B, 3000ft north of A. How fast are they separating 10 minutes after the second man starts?

Answer 1

Answer on Question #64137, Physics / Mechanics | Relativity

1. Two posts, one 8ft high and the other 12ft high, stand 15ft apart. They are to be stayed by wires attached to a single stake at ground level, the wires running to the tops of the posts. Where the stake should be placed, to use the least amount of wire?

Solution:

Therefore, using Pythagoras', the lengths of the wires are

$\vee (x^{2} + 8^{2})$ and $\vee ((15 - x)^{2} + 12^{2})$

Therefore the total length

$L = \vee (x^{2} + 64) + \vee (369 - 30x + x^{2})$

Differentiate using 'function of a function' (chain rule):

$L^{\prime} = x / \sqrt{(x^{2} + 64) + (x - 15)} /\sqrt{(369 - 30x + x^{2})}$

$\ldots = 0$ when

$x / \sqrt{(x^2 + 64) = -(x - 15)} / \sqrt{(369 - 30x + x^2)}$

Hence

$x \vee (369 - 30x + x^2) = (15 - x) \vee (x^2 + 64)$

Square both sides:

$x^{2}(369 - 30x + x^{2}) = (15 - x)^{2}(x^{2} + 64)$

$369x^{2} - 30x^{3} + x^{4} = (225 - 30x + x^{2})(x^{2} + 64)$

$369x^{2} - 30x^{3} + x^{4} = x^{4} - 30x^{3} + 289x^{2} - 1920x + 14400$

$80x^{2} + 1920x - 14400 = 0$

$x^{2} + 24x - 180 = 0$

$(x + 30)(x - 6) = 0$

Since $x > 0$

$x = 6$

Answer: 6

2. A man starts walking eastward at 5ft/sec from a point A. Ten minutes later a second man starts walking west at the rate of 5ft/sec from a point B, 3000ft north of A. How fast are they separating 10 minutes after the second man starts?

Solution:

$[dz/dt]_{t=1200}$

$\mathrm{dx} / \mathrm{dt} = 5$

$\mathrm{dy} / \mathrm{dt} = 5$

$z = \sqrt{(3000)^2 + (x + y)^2}$

$\mathrm{dz} / \mathrm{dt} = \frac{1}{2} [3000^2 + (x + y)^2]^{-1/2} 2(x + y)(\mathrm{dx} / \mathrm{dt} + \mathrm{dy} / \mathrm{dt})$

$t = 1200$

$x = 5$ (1200) $= 6000$

$y = 5(600) = 3000$

$$[dz/dt]_{t=1200} = [3000^2 + (6000 + 3000)^2]^{-1/2} \cdot 2 \cdot (6000 + 300) \cdot (5 + 5) = \sqrt{90} \, \text{ft/sec}$$

3. A man 6 feet tall walks away from the point under the light at the rate of 5 ft/sec. How fast is his shadow lengthening when he is 20 feet away from the point under the light?

**Solution:**

$$dx/dt = 5$$ $$20/6 = y / (y - x)$$ $$20y - 20x = 6y$$ $$14y = 20x$$ $$y = 20x / 14$$ $$y = 10x / 7$$ $$dy/dt = (10/7) \, dx/dt$$ $$dy/dt = (10/7) \cdot 5 = 50 / 5 \, \text{ft/sec}$$ $$d(y - X)/dt = dy/dt - dx/dt = (50 / 5) - 5 = 15/7 \, \text{ft/sec}$$

**Answer:** 15/7 ft/sec

4. At a given instant the legs of a right triangle are 8 in., and 6 in., respectively. The first leg decreases at 1 in/min and the second increases at 2 in/min. At what rate is the area increasing after 2 min.?

**Solution:**

$$A = 1/2 \, \text{bh}$$ $$db/dt = -1$$ $$dh/dt = 2$$

The attempt at a solution

$$A = 1/2 \, \text{bh}$$ $$dA/dt = 1/2 \cdot (db/dt * h + b * dh/dt)$$ $$dA/dt = 1/2 \cdot (-1 * 10 + 6 * 2)$$ $$dA/dt = 1/2 \cdot (2) = 1$$

**Answer:** 1

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