MATHEMATICS COLLEGE

A trough has a semicircular cross section with a radius of 9 feet. Water starts flowing into the trough in such a way that the depth of the water is increasing at a rate of 2 inches per hour. (a) Give a function w = f(t) relating the width w of the surface of the water to the time t, in hours. Make sure to specify the domain and compute the range too.(b) After how many hours will the surface of the water have width of 6 feet?

(c) Give a function t = f −^1 (w) relating the time to the width of the surface of the water. Make sure to specify the domain and compute the range too.

Answers

Answer 1

Answer:

Answer:

(a) Let h represents the height of water and w represents the width of the water,

Since, the depth of the water is increasing at a rate of 2 inches per hour,

So, after t hours,

The height of water, h(t) = 2t inches = t/6 ft,

( ∵ 1 foot = 12 inches ⇒ 1 inch = 1/12 ft )

Thus, the distance distance from the centre to the top of the water, d = 9 - h(t) ( see in the diagram )

$d=9-(t)/(6)$ ,

By the Pythagoras theorem,

$d^2 + ((w)/(2))^2 = 9^2$

$(9-(t)/(6))^2 +(w^2)/(4) = 81$

$(t^2)/(36)-(18t)/(6) + (w^2)/(4)=0$

$(t^2 - 108t + 9w^2)/(36)=0$

$t^2 - 108t + 9w^2 =0$

$9w^2 = 108t - t^2$

$w = (1)/(3)√(108t - t^2)$

Since, diameter of the semicircular cross section is 18 ft,

So, 0 ≤ w ≤ 18,

i.e Range = [0, 18]

Also, w will be defined if 108t - t² ≥ 0

⇒ (108 - t)t ≥ 0,

⇒ 0 ≤ t ≤ 108

i.e Domain = [0, 108]

(b) If w = 6,

$6 =(1)/(3)√(108t - t^2)$

$18 =√(108t-t^2)$

$324 = 108t - t^2$

$\implies t^2 - 108t+ 324=0$

By using quadratic formula,

$\implies t = 3.088\text{ or }t = 104.912$

Hence, After 3.1 hours or 104.9 hours will the surface of the water have width of 6 feet.

(c) $w = (1)/(3)√(108t- t^2)$

$\implies 3w = √(108t- t^2)$

$9w^2 = 108t - t^2$

$-9w^2 = -108t + t^2$

$-9w^2 + 2916 = 2916 - 108t + t^2$

$2916 - 9w^2 = (t - 108)^2$

$(t-108) = √(2916 - 9w^2)$

$t = √(2916 - 9w^2) + 108$

For 0 ≤ w ≤ 18,

0 ≤ t ≤ 108,

So, Domain = [0, 18]

Range = [0, 108]

Answer 2

Answer:

Final answer:

The width of the water's surface in a semicircular trough can be represented by the function w=t/3 and its domain is t ≥ 0 and the range is 0 ≤ w ≤ 6. To have a 6 feet wide surface, thus, it would take 18 hours. The inverse function is t=3w, with a domain of 0 ≤ w ≤ 6 and range of t ≥ 0.

Explanation:

Given that the depth of the water is increasing at a rate of 2 inches per hour in a semicircular trough, we can convert this rate to feet per hour by dividing by 12, getting an increase of 1/6 feet per hour.

(a) We can express the width of the surface of the water as a function of time. We consider the cross-section of the trough is a semicircle. So, the radius of the water's surface will be the height of water, and this height increases at 1/6 feet per hour. Therefore, the width of the surface of water, w=2r=2*1/6t=t/3. The domain of the function is t ≥ 0 and the range is 0 ≤ w ≤ 6.

(b) We set w=6 in the function w=t/3 and solve for t. We get t=3*6=18 hours.

(c) The inverse function of w=t/3 is t=3w. The domain of the inverse function is 0 ≤ w ≤ 6 and the range is t ≥ 0.

Learn more about Mathematical functions here:

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Step-by-step explanation:

Answer:

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You originally draw a design for an art contest on a 4 in. x 5 in. card. The second phase of the contest requires the drawing to be transferred to an 8.5 in x 11 in. standard sheet of paper and utilize as much of the space on the paper as possible. You determine that the largest size one of the dimensions of your drawing can be is 10.5 in. What is the length of the other dimension if the two drawings are similar? Type your exact answer in the blank without the units, and round to the nearest tenths.

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In a study of plant safety, it was found that the time it took for machine operators to react to a warning light was normally distributed with a mean 1 second and standard deviation 0.3 second. Suppose a warning light visible to 4 operators goes on. What is the probability that the average (mean) reaction time of the 4 operators exceeds 1.25 seconds?

Answers

Answer:

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Central Limit Theorem

The Central Limit Theorem estabilishes that, for a random variable X, with mean $\mu$ and standard deviation $\sigma$ , a large sample size can be approximated to a normal distribution with mean $\mu$ and standard deviation $(\sigma)/(√(n))$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = (X - \mu)/(\sigma)$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

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