A piece of elastic is attached to two nails on a flat board, with a button attached to the midpoint of the elastic. The nails are 5 cm apart. You stretch the elastic by pulling the button along the board in a direction that is perpendicular to the line between the nails.
A. Fnd an equation that relates the total length of the elastic x to the distance y that the button has moved.
B. You pull the button at a constant 3 cm/sec. Find the rate at which the length of the elastic is increasing when it is 12 cm long.
A.
x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)
B.
x^2 = 4y^2 + 25
=> 2x dx/dt = 8y dy/dt
=> dx/dt = 4 (y/x) dy/dt
When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
=> dx/dt = 4 (2.727/12) x 3 cm/s
= 2.727 cm/s.
you have two right triangles of height y
2.5^2 + y^2 = (x/2)^2
or
25 + 4y^2 = x^2
8y dy/dt = 2x dx/dt
when x = 12, y = 5.454
8(5.454) = 2(12) dx/dt
dx/dt = 1.818
A.
x = 2√[ y^2 + (2.5)^2 ] cm (using Pythagorus theorem)
B.
x^2 = 4y^2 + 25
=> 2x dx/dt = 8y dy/dt
=> dx/dt = 4 (y/x) dy/dt
When x = 12 cm, y = (1/2)√[ 144 - 25 ] = 2.727 cm
=> dx/dt = 4 (2.727/12) x 3 cm/s
= 2.727 cm/s.
Pythagorous Theorem,
z^2 = y^2 + x^2, where y = 2.5 cm and x is the legth of the pull.
substituting z^2 = 2.5^2 +x^2, z =sqrt(x^2 +6.25)
dz/dx (2z) = 2x
dz/dx = (x/z) ......... (a)
but dz/dt = dz/dx * dx/dt
Thus, dz/dt = (x/z) * dx/dt = x/z * 3
when z = 12 , y = 5, x= sqrt(144-25) = sqrt(119)
Thus dz/dt = sqrt(119)/12 * 3
= sqrt(119)/4 =10.91 /4 =2.73 cm/sec
Rate of increase in length of elastic = 2.73 cm/sec
To find an equation that relates the total length of the elastic x to the distance y that the button has moved, let's break down the problem:
A. When the elastic is stretched along the board, it forms a right-angled triangle with the line between the nails as the hypotenuse. Let's call the length of the elastic x and the distance the button has moved y.
Since the nails are 5 cm apart, the hypotenuse is the total length of the elastic. Therefore, we can use the Pythagorean theorem to relate x and y:
x^2 = (y^2) + (5^2)
Simplifying the equation, we get:
x^2 = y^2 + 25
This is the equation that relates the total length of the elastic x to the distance y that the button has moved.
B. To find the rate at which the length of the elastic is increasing when it is 12 cm long, we need to differentiate the equation we found in part A with respect to time (since we are given the rate of change of y with respect to time).
Differentiating both sides of the equation, we get:
2x(dx/dt) = 2y(dy/dt)
Rearranging the equation, we have:
(dx/dt) = (y/x) * (dy/dt)
Substituting the values given, we have:
(dx/dt) = (y/12) * (3 cm/sec)
Since we are looking for the rate at which the length of the elastic is increasing (dx/dt), we need to find the values of y and x when the length of the elastic is 12 cm.
From the equation we derived in part A, x^2 = y^2 + 25, substituting x = 12:
12^2 = y^2 + 25
144 - 25 = y^2
119 = y^2
Taking the positive square root, we have:
y = √119
Now, substituting the values y = √119 and x = 12 into the equation for (dx/dt), we have:
(dx/dt) = (√119 / 12) * (3 cm/sec)
Simplifying further, we get:
(dx/dt) = (3√119 / 12) cm/sec
Therefore, when the length of the elastic is 12 cm, the rate at which its length is increasing is (3√119 / 12) cm/sec.
First, let's sketch what we can derive geometrically.
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(A) Given we know those two triangles are right, we can relate the hypotenuse \(\frac12x\) to the altitude \(y\) and base \(\frac52\) using the Pythagorean theorem, which we can rearrange to yield an adequate relation:$$\left(\frac12x\right)^2=y^2+\left(\frac52\right)^2\\\frac14x^2=y^2+\frac{25}4\\x^2=4y^2+25$$
(B) We're given that the button is moving at a rate of 3 cm/s, which can be expressed using a time derivative as \(\frac{dy}{dt}=3\). We're told that the elastic (at the instant we're interested in) is 12 cm long, i.e. \(x=12\); given this, we can determine the distance of the button from its initial position with relative ease:$$(12)^2=4y^2+25\\4y^2=144-25=119\\y^2=\frac{119}{4}\\y=\frac{\sqrt{119}}2$$Let's use implicit differentiation on our formula above to relate the rates of elongation:$$2x\frac{dx}{dt}=8y\frac{dy}{dt}\\24\frac{dx}{dt}=12\sqrt{119}\\2\frac{dx}{dt}=\sqrt{119}\\\frac{dx}{dt}=\frac{\sqrt{119}}2\approx5.4544$$
each side of the elastic is the hypotenuse of a triangle with legs 5/2 and y, so
x = 2√(2.5^2 + y^2)
dx/dt = 2y/√(2.5^2+y^2) dy/dt
when x=12, y=5.45, so
dx/dt = 2(5.45)/6 (-3) = -1/5.45 = -0.18 cm/s