the quadratic function h (t)=-16t^2+150 models a balls height, in feet, over time, in seconds, after it is dropped from a 15 sto
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Step-by-step explanation:
From the statement:
M: is total to be memorized
A(t): the amount memorized.
The key issue is translate this statement as equation "rate at which a subject is memorized is assumed to be proportional to the amount that is left to be memorized"
memorizing rate is .
the amount that is left to be memorized can be expressed as the total minus the amount memorized, that is .
So we can write
And that would be the differential equation for A(t).
That'd be true only if the value of "s" is the exact same one for both
namely if sec(s) = cos(s)
then solving for "s"
thus
![\bf sec(s)=cos(s)\qquad but\implies sec(\theta)=\cfrac{1}{cos(\theta)} \\\\\\ thus\cfrac{1}{cos(s)}=cos(s)\implies 1=cos^2(s)\implies \pm \sqrt{1}=cos(s) \\\\\\ \pm 1=cos(s)\impliedby \textit{now taking }cos^{-1}\textit{ to both sides} \\\\\\ cos^{-1}(\pm 1)=cos^{-1}[cos(s)]\implies cos^{-1}(\pm 1)=\measuredangle s](https://tex.z-dn.net/?f=%5Cbf%20sec%28s%29%3Dcos%28s%29%5Cqquad%20but%5Cimplies%20sec%28%5Ctheta%29%3D%5Ccfrac%7B1%7D%7Bcos%28%5Ctheta%29%7D%0A%5C%5C%5C%5C%5C%5C%0Athus%5Ccfrac%7B1%7D%7Bcos%28s%29%7D%3Dcos%28s%29%5Cimplies%201%3Dcos%5E2%28s%29%5Cimplies%20%5Cpm%20%5Csqrt%7B1%7D%3Dcos%28s%29%0A%5C%5C%5C%5C%5C%5C%0A%5Cpm%201%3Dcos%28s%29%5Cimpliedby%20%5Ctextit%7Bnow%20taking%20%7Dcos%5E%7B-1%7D%5Ctextit%7B%20to%20both%20sides%7D%0A%5C%5C%5C%5C%5C%5C%0Acos%5E%7B-1%7D%28%5Cpm%201%29%3Dcos%5E%7B-1%7D%5Bcos%28s%29%5D%5Cimplies%20cos%5E%7B-1%7D%28%5Cpm%201%29%3D%5Cmeasuredangle%20s)
namely if sec(s) = cos(s)
then solving for "s"
thus
We're given the implicit equation
x² + 2x + y² - 4y = 12
where "implicit" means that one of the variables is considered a function of the other. In this case, the variable y = y(x) depends on the value of the variable x - in other words, the possible values of y are implied by the value of x. We then say that this equation defines y as an implicit function of x.
We're also given the implicit derivative,
dy/dx = (-2x - 2)/(2y - 4)
which tells us the rate of change y as a function of both x and y. The value of dy/dx corresponds to the slope of the tangent line to the curve defined by the implicit equation above at some point (x, y).
The tangent line is horizontal when its slope is zero. This happens when
(-2x - 2)/(2y - 4) = 0
If y ≠ 2, then we can eliminate the denominator and we're left with
-2x - 2 = 0
Solve for x :
-2x = 2
x = -1
This tells us that the points on the curve with x-coordinate -1 have a tangent line to the curve that is horizontal.
The tangent line is vertical when its slope is undefined/infinite, or equivalently when the denominator of dy/dx is zero:
2y - 4 = 0
Solve for y :
2y = 4
y = 2
So any point on the curve with y-coordinate 2 will have a tangent line there that is vertical.
By completing the square in the implicit equation, we can easily identify where these points are located.
x² + 2x + y² - 4y = 12
x² + 2x + 1 + y² - 4y + 4 = 17
(x + 1)² + (y - 2)² = 17
This equation defines a circle centered at (-1, 2) with radius √17. When x = -1, we have
(y - 2)² = 17 ⇒ y = 2 ± √17
and when y = 2, we have
(x + 1)² = 17 ⇒ x = -1 ± √17
See the attached plot to see the circle and the tangents at these points.
