Solved on Dec 16, 2023

Find the speed of the tip of a woman's shadow when she is $15 \mathrm{~m}$ from a $3 \mathrm{~m}$ tall pole, running at $1.8 \mathrm{~m} / \mathrm{s}$ and is $1.5 \mathrm{~m}$ tall.

STEP 1

Assumptions
1. The streetlight is at the top of a pole that is $3 \text{ m}$ tall.
2. The woman is $1.5 \text{ m}$ tall.
3. The woman runs away from the pole at a speed of $1.8 \text{ m/s}$ .
4. The woman is running along a straight path.
5. We are asked to find the speed of the tip of her shadow when she is $15 \text{ m}$ from the base of the pole.
6. We assume that the light source (streetlight) is a point source at the top of the pole.

STEP 2

Let's denote the distance of the woman from the pole as $x(t)$ , where $t$ is time. The length of the shadow on the ground is $y(t)$ , and the distance from the top of the pole to the tip of the shadow is $z(t)$ . The woman's height is $h_w = 1.5 \text{ m}$ , and the pole's height is $h_p = 3 \text{ m}$ .

STEP 3

We can use similar triangles to relate the heights and distances involved. The ratio of the woman's height to the pole's height is equal to the ratio of the distance of the woman from the pole to the distance from the top of the pole to the tip of her shadow.
$\frac{h_w}{h_p} = \frac{x(t)}{z(t)}$

STEP 4

From the similar triangles, we can express $z(t)$ in terms of $x(t)$ .
$z(t) = \frac{h_p}{h_w} \cdot x(t)$

STEP 5

We know that $y(t)$ , the length of the shadow, is the difference between $z(t)$ and $x(t)$ .
$y(t) = z(t) - x(t)$

STEP 6

Substitute the expression for $z(t)$ from STEP_4 into the equation for $y(t)$ .
$y(t) = \frac{h_p}{h_w} \cdot x(t) - x(t)$

STEP 7

Factor out $x(t)$ from the right side of the equation.
$y(t) = x(t) \left(\frac{h_p}{h_w} - 1\right)$

STEP 8

Now, we differentiate both sides of the equation with respect to time $t$ to find the rate at which the tip of the shadow is moving, which is $\frac{dy}{dt}$ .
$\frac{d}{dt} y(t) = \frac{d}{dt} \left[ x(t) \left(\frac{h_p}{h_w} - 1\right) \right]$

STEP 9

Apply the product rule to the right side of the equation, noting that $\left(\frac{h_p}{h_w} - 1\right)$ is a constant.
$\frac{dy}{dt} = \left(\frac{h_p}{h_w} - 1\right) \frac{dx}{dt}$

STEP 10

We know the speed of the woman, which is the rate of change of her distance from the pole, $\frac{dx}{dt} = 1.8 \text{ m/s}$ .

STEP 11

Substitute the known values for $h_p$ , $h_w$ , and $\frac{dx}{dt}$ into the equation.
$\frac{dy}{dt} = \left(\frac{3}{1.5} - 1\right) \cdot 1.8$

STEP 12

Calculate the constant factor.
$\frac{3}{1.5} - 1 = 2 - 1 = 1$

STEP 13

Now, calculate the rate at which the tip of the shadow is moving, $\frac{dy}{dt}$ .
$\frac{dy}{dt} = 1 \cdot 1.8 = 1.8 \text{ m/s}$
The tip of the woman's shadow is moving at $1.8 \text{ m/s}$ when she is $15 \text{ m}$ from the base of the pole.

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Find the speed of the tip of a woman's shadow when she is $15 \mathrm{~m}$ from a $3 \mathrm{~m}$ tall pole, running at $1.8 \mathrm{~m} / \mathrm{s}$ and is $1.5 \mathrm{~m}$ tall.

STEP 1

STEP 2

STEP 3

We can use similar triangles to relate the heights and distances involved. The ratio of the woman's height to the pole's height is equal to the ratio of the distance of the woman from the pole to the distance from the top of the pole to the tip of her shadow.
$\frac{h_w}{h_p} = \frac{x(t)}{z(t)}$

STEP 4

From the similar triangles, we can express $z(t)$ in terms of $x(t)$ .
$z(t) = \frac{h_p}{h_w} \cdot x(t)$

STEP 5

We know that $y(t)$ , the length of the shadow, is the difference between $z(t)$ and $x(t)$ .
$y(t) = z(t) - x(t)$

STEP 6

Substitute the expression for $z(t)$ from STEP_4 into the equation for $y(t)$ .
$y(t) = \frac{h_p}{h_w} \cdot x(t) - x(t)$

STEP 7

Factor out $x(t)$ from the right side of the equation.
$y(t) = x(t) \left(\frac{h_p}{h_w} - 1\right)$

STEP 8

Now, we differentiate both sides of the equation with respect to time $t$ to find the rate at which the tip of the shadow is moving, which is $\frac{dy}{dt}$ .
$\frac{d}{dt} y(t) = \frac{d}{dt} \left[ x(t) \left(\frac{h_p}{h_w} - 1\right) \right]$

STEP 9

Apply the product rule to the right side of the equation, noting that $\left(\frac{h_p}{h_w} - 1\right)$ is a constant.
$\frac{dy}{dt} = \left(\frac{h_p}{h_w} - 1\right) \frac{dx}{dt}$

STEP 10

We know the speed of the woman, which is the rate of change of her distance from the pole, $\frac{dx}{dt} = 1.8 \text{ m/s}$ .

STEP 11

Substitute the known values for $h_p$ , $h_w$ , and $\frac{dx}{dt}$ into the equation.
$\frac{dy}{dt} = \left(\frac{3}{1.5} - 1\right) \cdot 1.8$

STEP 12

Calculate the constant factor.
$\frac{3}{1.5} - 1 = 2 - 1 = 1$

STEP 13

Now, calculate the rate at which the tip of the shadow is moving, $\frac{dy}{dt}$ .
$\frac{dy}{dt} = 1 \cdot 1.8 = 1.8 \text{ m/s}$
The tip of the woman's shadow is moving at $1.8 \text{ m/s}$ when she is $15 \text{ m}$ from the base of the pole.

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Find the speed of the tip of a woman's shadow when she is 15 m15 \mathrm{~m}15 m from a 3 m3 \mathrm{~m}3 m tall pole, running at 1.8 m/s1.8 \mathrm{~m} / \mathrm{s}1.8 m/s and is 1.5 m1.5 \mathrm{~m}1.5 m tall.

STEP 1

STEP 2

STEP 3

STEP 4

From the similar triangles, we can express z(t)z(t)z(t) in terms of x(t)x(t)x(t).z(t)=hphw⋅x(t)z(t) = \frac{h_p}{h_w} \cdot x(t)z(t)=hw​hp​​⋅x(t)

STEP 5

We know that y(t)y(t)y(t), the length of the shadow, is the difference between z(t)z(t)z(t) and x(t)x(t)x(t).y(t)=z(t)−x(t)y(t) = z(t) - x(t)y(t)=z(t)−x(t)

STEP 6

Substitute the expression for z(t)z(t)z(t) from STEP_4 into the equation for y(t)y(t)y(t).y(t)=hphw⋅x(t)−x(t)y(t) = \frac{h_p}{h_w} \cdot x(t) - x(t)y(t)=hw​hp​​⋅x(t)−x(t)

STEP 7

Factor out x(t)x(t)x(t) from the right side of the equation.y(t)=x(t)(hphw−1)y(t) = x(t) \left(\frac{h_p}{h_w} - 1\right)y(t)=x(t)(hw​hp​​−1)

STEP 8

STEP 9

Apply the product rule to the right side of the equation, noting that (hphw−1)\left(\frac{h_p}{h_w} - 1\right)(hw​hp​​−1) is a constant.dydt=(hphw−1)dxdt\frac{dy}{dt} = \left(\frac{h_p}{h_w} - 1\right) \frac{dx}{dt}dtdy​=(hw​hp​​−1)dtdx​

STEP 10

We know the speed of the woman, which is the rate of change of her distance from the pole, dxdt=1.8 m/s\frac{dx}{dt} = 1.8 \text{ m/s}dtdx​=1.8 m/s.

STEP 11

Substitute the known values for hph_php​, hwh_whw​, and dxdt\frac{dx}{dt}dtdx​ into the equation.dydt=(31.5−1)⋅1.8\frac{dy}{dt} = \left(\frac{3}{1.5} - 1\right) \cdot 1.8dtdy​=(1.53​−1)⋅1.8

STEP 12

Calculate the constant factor.31.5−1=2−1=1\frac{3}{1.5} - 1 = 2 - 1 = 11.53​−1=2−1=1

STEP 13

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Find the speed of the tip of a woman's shadow when she is $15 \mathrm{~m}$ from a $3 \mathrm{~m}$ tall pole, running at $1.8 \mathrm{~m} / \mathrm{s}$ and is $1.5 \mathrm{~m}$ tall.

From the similar triangles, we can express $z(t)$ in terms of $x(t)$ .
$z(t) = \frac{h_p}{h_w} \cdot x(t)$

We know that $y(t)$ , the length of the shadow, is the difference between $z(t)$ and $x(t)$ .
$y(t) = z(t) - x(t)$

Substitute the expression for $z(t)$ from STEP_4 into the equation for $y(t)$ .
$y(t) = \frac{h_p}{h_w} \cdot x(t) - x(t)$

Factor out $x(t)$ from the right side of the equation.
$y(t) = x(t) \left(\frac{h_p}{h_w} - 1\right)$

Apply the product rule to the right side of the equation, noting that $\left(\frac{h_p}{h_w} - 1\right)$ is a constant.
$\frac{dy}{dt} = \left(\frac{h_p}{h_w} - 1\right) \frac{dx}{dt}$

We know the speed of the woman, which is the rate of change of her distance from the pole, $\frac{dx}{dt} = 1.8 \text{ m/s}$ .

Substitute the known values for $h_p$ , $h_w$ , and $\frac{dx}{dt}$ into the equation.
$\frac{dy}{dt} = \left(\frac{3}{1.5} - 1\right) \cdot 1.8$

Calculate the constant factor.
$\frac{3}{1.5} - 1 = 2 - 1 = 1$