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3 years ago

Find the surface area

Mathematics

1 answer:

mixas84 [53]3 years ago

7 0

Surface Area formula of a cylinder:
S.A = 2 × Area of base + Perimeter of base × Height of prism
= 2 $\pi$ r² + 2 $\pi$ rh
= 2 $\pi$ (5)² + 2 $\pi$ (5)(18)
= 50 $\pi$ + 180 $\pi$
= 230 $\pi$ in² ( in terms of $\pi$ )
= 722.566 in²
= 722.7 in² ( to the nearest tenth )

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3 years ago

A ladder 16 feet long is leaning against the wall of a tall building. The base of the ladder is moving away from the wall at a r

svet-max [94.6K]

Answer:

a. 0.588

b. 0.0722

c. 4.576 sqft/sec

Step-by-step explanation:

Let b and h denote the base and height as indicated in the diagram. By pythagoras theorem, $h^2 + b^2 = 16^2 = 256 \dotsc\;(1)$ because it is a right angle triangle.

It is given that $\frac{db}{dt} = 1$

Now differentiate (1) with respect to t (time) :

$\displaystyle{2h\frac{dh}{dt} + 2b\frac{db}{dt} = 0 \implies \frac{dh}{dt} = -\frac{b}{h} \frac{db}{dt}}$

$\displaystyle{=-\frac{b}{\sqrt{256 - b^2}} \frac{db}{dt} = -\frac{8}{13.856} \times 1 = -0.588}$

The minus sign indicates that the value of h is actually decreasing. The required answer is 0.588.

b. From the diagram, infer that $16 \sin{\theta} = b$ . When b = 8, then $\theta = \arcsin{0.5} = \ang{30}$ .

Differentiate the above equation w.r.t t

$\displaystyle{16 \cos{\theta} \frac{d\theta}{dt} = \frac{db}{dt} \implies \frac{d\theta}{dt} = \frac{1}{16 \cos{\theta}} = \frac{1}{13.856} = \mathbf{0.0722}}$

c. The area of the triangle is given by $A = 0.5\times h \times b$ . Differentiating w.r.t t,

$\displatstyle{\frac{dA}{dt} = 0.5 b \frac{dh}{dt} + 0.5 h \frac{db}{dt}}$

Plugging in b = 8, h = 13.856, $\frac{dh}{dt} = -0.588$ ,

$\frac{dA}{dt} = -2.352 + 6.928 = \mathbf{4.576 ft^2/sec}$

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