A man starts at a point A and walks 18 feet north. He then turns and walks due east at 18 feet per second. If a searchlight placed at A follows him, at what rate is the light turning 3 seconds after he started walking east

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Answer:

1/10 per sec

Step-by-step explanation:

When he's walked x feet in the eastward direction, the angle Θ that the search light makes has tangent

tanΘ = x/18

Taking the derivative with respect to time

sec²Θ dΘ/dt = 1/18 dx/dt.

He's walking at a rate of 18 ft/sec, so dx/dt = 18.

After 3seconds,

Speed = distance/time

18ft/sec =distance/3secs

x = 18 ft/sec (3 sec)

= 54ft. At this moment

tanΘ = 54/18

= 3

sec²Θ = 1 + tan²Θ

1 + 3² = 1+9

= 10

So at this moment

10 dΘ/dt = (1/18ft) 18 ft/sec = 1

10dΘ/dt = 1

dΘ/dt = 1/10 per sec

At 1/10 per second is the rate when the light turns 3 seconds after he started walking east.

It is given that a man starts at point A and walks 18 feet north. He then turns and walks due east at 18 feet per second.

It is required to find at what rate is the light turning 3 seconds after he started walking east.

What is the trigonometric ratio?

The trigonometric ratio is defined as the ratio of the pair of a right-angled triangle.

The [tex]\rm tan\theta[/tex] is the ratio of the perpendicular to the base.

When he started walking eastward direction, the searchlight makes an angle [tex]\theta[/tex]

Let the distance is x

And the  [tex]\rm tan\theta = \frac{x}{18}[/tex]

After performing the derivate with respect to time, we get:

[tex]\rm sec^2\theta\frac{d\theta}{dt} = \frac{1}{18} \frac{dx}{dt}[/tex] ...(1)    ( the differentiation of [tex]\rm tan\theta \ is \ sec^2\theta\\[/tex])

He walking at the rate of 18 ft per second ie.

[tex]\rm \frac{dx}{dt} = 18[/tex]  

After 3 seconds  [tex]\rm Speed=\frac{Distance}{Time}[/tex]

By using speed- time formula we can calculate the distance:

Distance x = 18×3 ⇒ 54 ft.

[tex]\rm tan\theta = \frac{54}{18}[/tex]  ⇒3

We know that:

[tex]\rm sec^2\theta = 1+tan^2\theta[/tex]

[tex]\rm sec^2\theta = 1+3^2\\\\\rm sec^2\theta = 10\\[/tex]

Put this value in the (1) equation, we get:

[tex]\rm 10\frac{d\theta}{dt} = \frac{1}{18} \times18[/tex]    ∵ [tex]\rm (\frac{dx}{dt} =18)[/tex]

[tex]\rm 10\frac{d\theta}{dt} = 1\\\\\rm \frac{d\theta}{dt} = \frac{1}{10} \\\\[/tex]per second.

Thus, at 1/10 per second what rate is the light turning 3 seconds after he started walking east.

Know more about trigonometry here:

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