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dexar [7]
3 years ago
10

A ball thrown in the air has a height of y = - 16x² + 50x + 3 feet after x seconds. a) What are the units of measurement for the

rate of change of y? b) Find the rate of change of y between x = 0 and x = 2?
Mathematics
1 answer:
Lubov Fominskaja [6]3 years ago
3 0
<h2>Answer:</h2>

(a) ft/s

(b) 1ft/s

<h2>Step-by-step explanation:</h2>

Given equation;

y = (- 16x² + 50x + 3)ft       -------------(i)

Where;

y is measured in feet(ft)

x is measured in seconds(s).

(a) The rate of change of y with respect to x is found by dividing the total change in y by the total change in x. i.e

Δy / Δx

Where;

Δy = y₂ - y₁  

Δx = x₂ - x₁

∴ Δy / Δx = \frac{y_2 - y_1}{x_2 - x_1}              --------------(ii)

Since y is measured in feet, Δy will also be measured in feet.

Also, since x is measured in seconds, Δx will also be measured in seconds.

Therefore, the rate of change of y with respect to x (Δy / Δx) will be measured in feet per second (ft/s)

(b) The rate of change of y between x  = 0 and x = 2 can be found by using equation (ii)

Where;

y₂ is the value of y at x = 2 found by substituting x = 2 into equation (i)

y₁ is the value of y at x = 0 found by substituting x = 0 into equation (i)

=> y₂ =  - 16(2)² + 50(2) + 3 = 39

=> y₁ =  - 16(1)² + 50(1) + 3 = 37

Now, substitute the values of y₂, y₁, x₂ and x₁  into equation (ii)

Δy / Δx = \frac{39 - 37}{2 - 0}  

Δy / Δx = \frac{2}{2}

Δy / Δx = 1

Therefore, the rate of change of y is 1 ft/s

 

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Use Euler's method with step size 0.2 to estimate y(1), where y(x) is the solution of the initial-value problem y' = x2y − 1 2 y
irina [24]

Answer:

Therefore the value of y(1)= 0.9152.

Step-by-step explanation:

According to the Euler's method

y(x+h)≈ y(x) + hy'(x) ....(1)

Given that y(0) =3 and step size (h) = 0.2.

y'(x)= x^2y(x)-\frac12y^2(x)

Putting the value of y'(x) in equation (1)

y(x+h)\approx y(x) +h(x^2y(x)-\frac12y^2(x))

Substituting x =0 and h= 0.2

y(0+0.2)\approx y(0)+0.2[0\times y(0)-\frac12 (y(0))^2]

\Rightarrow y(0.2)\approx 3+0.2[-\frac12 \times3]    [∵ y(0) =3 ]

\Rightarrow y(0.2)\approx 2.7

Substituting x =0.2 and h= 0.2

y(0.2+0.2)\approx y(0.2)+0.2[(0.2)^2\times y(0.2)-\frac12 (y(0.2))^2]

\Rightarrow y(0.4)\approx  2.7+0.2[(0.2)^2\times 2.7- \frac12(2.7)^2]

\Rightarrow y(0.4)\approx 1.9926

Substituting x =0.4 and h= 0.2

y(0.4+0.2)\approx y(0.4)+0.2[(0.4)^2\times y(0.4)-\frac12 (y(0.4))^2]

\Rightarrow y(0.6)\approx  1.9926+0.2[(0.4)^2\times 1.9926- \frac12(1.9926)^2]

\Rightarrow y(0.6)\approx 1.6593

Substituting x =0.6 and h= 0.2

y(0.6+0.2)\approx y(0.6)+0.2[(0.6)^2\times y(0.6)-\frac12 (y(0.6))^2]

\Rightarrow y(0.8)\approx  1.6593+0.2[(0.6)^2\times 1.6593- \frac12(1.6593)^2]

\Rightarrow y(0.6)\approx 0.8800

Substituting x =0.8 and h= 0.2

y(0.8+0.2)\approx y(0.8)+0.2[(0.8)^2\times y(0.8)-\frac12 (y(0.8))^2]

\Rightarrow y(1.0)\approx  0.8800+0.2[(0.8)^2\times 0.8800- \frac12(0.8800)^2]

\Rightarrow y(1.0)\approx 0.9152

Therefore the value of y(1)= 0.9152.

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