find dy/dx and (d^2)y/(d^2)x for x=2sint y=3cost 0<1<2pi
dy/dx = y'/x' = (-3sint)/(2cost) = -3/2 tan t
d^2y/dx^2 = (x'y"-x"y')/x'^3
= ((2cost)(-3cost) - (-2sint)(-3sint))/(8cos^3 t)
= -3/4 sec^3 t
Check:
y = 3/2 √(4-x^2)
y' = -3/2 x/√(4-x^2)
y" = -6/(4-x^2)3/2
I'll leave it to you to verify that they are the same.
To find dy/dx, we need to differentiate y with respect to x. Given that y = 3cos(t), where x = 2sin(t), we can substitute for t using the relationship x = 2sin(t).
To do so, we'll use the Chain Rule. The Chain Rule states that dy/dx = dy/dt * dt/dx.
First, let's find dy/dt.
dy/dt = -3sin(t)
Next, let's find dt/dx.
dt/dx = 1/(dx/dt)
= 1/(d(2sin(t))/dt)
= 1/(2cos(t))
= 1/(2cos(arcsin(x/2)))
= 1/(2√(4-x^2/4))
= 1/(2√(16-x^2))/2
= 1/√(16-x^2)
Now, let's substitute dy/dt and dt/dx back into the Chain Rule equation:
dy/dx = (-3sin(t)) / (√(16-x^2))
Finally, we'll evaluate dy/dx at x = 2sin(t):
dy/dx = (-3sin(t)) / (√(16-(2sin(t))^2))
= (-3sin(t)) / (√(16-(4sin(t))^2))
= (-3sin(t)) / (√(16-16sin^2(t)))
= (-3sin(t)) / (√(16(1-sin^2(t))))
= (-3sin(t)) / (4cos(t))
= -3sin(t) / 4cos(t)
= -3tan(t) / 4
To find the second derivative, (d^2)y/(d^2)x, we'll differentiate dy/dx with respect to x.
Let's differentiate dy/dx with respect to x:
(d^2)y / (d^2)x = d/dx(-3tan(t)/4)
Since tan(t) is equivalent to sin(t)/cos(t), we can rewrite the equation as follows:
(d^2)y / (d^2)x = d/dx(-3sin(t)/4cos(t))
Applying the Quotient Rule, we have:
(d^2)y / (d^2)x = [(-3cos(t)(d(sin(t))/dx)) - (-3sin(t)(d(cos(t))/dx))] / (4cos^2(t))
Let's evaluate d(sin(t))/dx and d(cos(t))/dx:
d(sin(t))/dx = d(sin(t))/dt * dt/dx
= cos(t) * (1/√(16-x^2))
d(cos(t))/dx = d(cos(t))/dt * dt/dx
= -sin(t) * (1/√(16-x^2))
Substituting these values back into the equation, we get:
(d^2)y / (d^2)x = [(-3cos(t)(cos(t) * (1/√(16-x^2)))) - (-3sin(t)(-sin(t) * (1/√(16-x^2))))] / (4cos^2(t))
Simplifying further, we have:
(d^2)y / (d^2)x = [(-3cos^2(t)/(√(16-x^2))) - (3sin^2(t)/(√(16-x^2)))] / (4cos^2(t))
Substituting x = 2sin(t) and x = 2cos(t), we can eliminate trigonometric functions and simplify our expression further. However, since you only provided the range of t (0 < t < 2π), and not the actual value, we cannot compute the second derivative accurately for this given range.
To find dy/dx, we need to differentiate y with respect to x. But in order to do that, we first need to express y in terms of x instead of t.
Given:
x = 2sin(t)
y = 3cos(t)
We can solve for t in terms of x, and then substitute it into the equation for y.
x = 2sin(t)
Dividing both sides by 2:
x/2 = sin(t)
Since the range of sin(t) is [-1, 1], we can write:
-1 ≤ sin(t) ≤ 1
Now, to find the value of t, we can take the inverse sine (arcsine) of both sides:
arcsin(-1) ≤ arcsin(sin(t)) ≤ arcsin(1)
-π/2 ≤ t ≤ π/2
Since we know that t is between 0 and 2π (0 ≤ t ≤ 2π), we can see that the value of t must be between π/2 and 3π/2.
To determine the quadrant where t lies, we use the property of the cosine function, which is positive in the first and fourth quadrants:
cos(t) > 0 when π/2 ≤ t ≤ 3π/2
Therefore, the value of t lies in the interval [π/2, 3π/2].
Now, we can rewrite y in terms of x by substituting the value of t we found:
y = 3cos(t)
y = 3cos(arcsin(x/2))
y = 3√(1 - (x/2)^2)
Now, we can differentiate y with respect to x by applying the chain rule.
dy/dx = d/dx [3√(1 - (x/2)^2)]
Using the chain rule, we have:
dy/dx = 3 * (1/2) * (-2x/4) * √(1 - (x/2)^2)^(-1/2)
dy/dx = -3x/4√(1 - (x/2)^2)
Therefore, the derivative dy/dx is equal to -3x/4√(1 - (x/2)^2).
To find (d^2)y/(d^2)x, we need to differentiate dy/dx again:
(d^2)y/(d^2)x = d/dx [-3x/4√(1 - (x/2)^2)]
To differentiate this expression, we need to use the product rule and the chain rule.
(d^2)y/(d^2)x = [-3(1/4√(1 - (x/2)^2)) - 3x * (d/dx(1/4√(1 - (x/2)^2)))]
Now, we can compute the derivative of the second term using the chain rule:
(d/dx(1/4√(1 - (x/2)^2))) = (1/4) * (-1/2) * (2x/4) * √(1 - (x/2)^2)^(-1/2)
(d/dx(1/4√(1 - (x/2)^2))) = -x/(8√(1 - (x/2)^2))
Substituting this back into the equation, we get:
(d^2)y/(d^2)x = [-3(1/4√(1 - (x/2)^2)) - 3x * (-x/(8√(1 - (x/2)^2)))]
Simplifying further, we have:
(d^2)y/(d^2)x = -3/(4√(1 - (x/2)^2)) - 3x^2/(8(1 - (x/2)^2)√(1 - (x/2)^2))
Therefore, the second derivative (d^2)y/(d^2)x is equal to -3/(4√(1 - (x/2)^2)) - 3x^2/(8(1 - (x/2)^2)√(1 - (x/2)^2)).