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3 years ago

15

given the height and the circumference of the base of a cylinder describe how to find the surface area of the entire cylinder

1 answer:

Dafna11 [192]3 years ago

7 0

The radius of the cylinder is related to its circumference by
.. C = 2πr
.. r = C/(2π)
The area of the cylinder base is given by
.. A = π*r^2 = π*(C/(2π))^2 = C^2/(4π)

The lateral area of the cylinder is given by
.. S = C*h

The total area of the cylinder is the lateral area plus the area of both bases.
.. total area = C*h +2*C^2/(4π)
.. total area = C*h +C^2/(2π) . . . . . . . . area from circumference and height

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Let h(x)=20e^kx where k ɛ R (Picture attached. Thank you so much!)

zloy xaker [14]

Answer:

A)

$k=0$

B)

$\displaystyle \begin{aligned} 2k + 1& = 2\ln 20 + 1 \\ &\approx 2.3863\end{aligned}$

C)

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Step-by-step explanation:

We are given the function:

$\displaystyle h(x) = 20e^{kx} \text{ where } k \in \mathbb{R}$

A)

Given that h(1) = 20, we want to find <em>k</em>.

h(1) = 20 means that <em>h</em>(x) = 20 when <em>x</em> = 1. Substitute:

$\displaystyle (20) = 20e^{k(1)}$

Simplify:

$1= e^k$

Anything raised to zero (except for zero) is one. Therefore:

$k=0$

B)

Given that h(1) = 40, we want to find 2<em>k</em> + 1.

Likewise, this means that <em>h</em>(x) = 40 when <em>x</em> = 1. Substitute:

$\displaystyle (40) = 20e^{k(1)}$

Simplify:

$\displaystyle 2 = e^{k}$

We can take the natural log of both sides:

$\displaystyle \ln 2 = \underbrace{k\ln e}_{\ln a^b = b\ln a}$

By definition, ln(e) = 1. Hence:

$\displaystyle k = \ln 2$

Therefore:

$2k+1 = 2\ln 2+ 1 \approx 2.3863$

C)

Given that h(1) = 10, we want to find <em>k</em> - 3.

Again, this meas that <em>h</em>(x) = 10 when <em>x</em> = 1. Substitute:

$\displaystyle (10) = 20e^{k(1)}$

Simplfy:

$\displaystyle \frac{1}{2} = e^k$

Take the natural log of both sides:

$\displaystyle \ln \frac{1}{2} = k\ln e$

Therefore:

$\displaystyle k = \ln \frac{1}{2}$

Therefore:

$\displaystyle k - 3 = \ln\frac{1}{2} - 3\approx-3.6931$

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