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3 years ago

A 12-foot wire supports a tree creating a 48 degreee angle. What is the closest distance up the tree that the wire reaches?

Mathematics

1 answer:

BartSMP [9]3 years ago

5 0

The closest distance up the tree where the wire reaches is 17.93 foot.

<u>Explanation:</u>

Given-

Length of the wire, L = 12 foot

angle, Ф= 48°

Distance upto which the wire reaches the tree, x = ?

We know,

cos Ф = base/ hypotenuse

cos 48° = 12/ x

0.669 = 12/x

x = 17.93 foot

Therefore, the closest distance up the tree where the wire reaches is 17.93 foot.

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You are situated 300 feet from the base of Tower Glitz Plaza watching an external elevator descend down the side of the building

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Answer:

18.66 ft/s

Step-by-step explanation:

The distance between you and the elevator is given by:

$h=\sqrt{x^2+y^2}$

The rate of change for the distance between you and the elevator is given by:

$\frac{dh}{dt}=\frac{dh}{dy}*\frac{dy}{dt}$

$-16=\frac{dh}{dy}*\frac{dy}{dt}$

$\frac{dh}{dy}=\frac{d}{dy} (\sqrt{x^2+y^2})\\$

Applying the chain rule:

$u=x^2+y^2\\\frac{dh}{dy}=\frac{d\sqrt u}{du} *\frac{du}{dy}\\\frac{dh}{dy}=\frac{1}{2\sqrt u} *2y\\\frac{dh}{dy}=\frac{y}{\sqrt {(x^2+y^2)}}$

Therefore, at x=300 and y = 500, dy/dt is:

$-16=\frac{y}{\sqrt {(x^2+y^2)}}*\frac{dy}{dt}\\-16=\frac{500}{\sqrt {(300^2+500^2)}}*\frac{dy}{dt}\\\frac{dy}{dt}=-18.66\ ft/s$

The elevator is descending at 18.66 ft/s.

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