1.) which of the following represents dy/dx when y=e^-2x Sec(3x)?
A.3e^-2x sec(3x) tan (3x)-2e^-2x Sec(3x)<<<< my choice
B.)3e^-2x sec(3x)tan (3x)-2xe^-2x sec(3x)
C.)3e^-2x sec(3x)tan (x)-2e^-2x Sec(3x)
D.)3e^-2x sec(3x)tan (x)-2xe^-2x sec(3x)
2.)calculate dy/dx if y=Ln(2x^3+3x)
A.)1/2x^3+3x
B.)1/6x^2+3
C.)2x^3+3x/6x^2+3
D.)6x^2+3/2x^3+3x <<<< my choice
3.)Given the curve that is described by the equation r=3 cos theta, find the angle that tangent lines makes with the radius vector when theta=120.
A.) 30deg
B.) 45deg
C.) 60deg
D.) 90deg << my choice
4.)A line rotates in a horizontal plane according to the equation theta=2^t-6t,where theta is the angular position of the rotating line, in radians ,and tis the time,in seconds. Determine the angular acceleration when t=2sec.
A.) 6 radians per sec^2
B.) 12 radians per sec^2
C.) 18 radians per sec^2
D.) 24 radians per sec^2 << my choice.
1.) To find dy/dx, we need to apply the product and chain rule in calculus.
First, let's break down the given function: y = e^-2x * Sec(3x).
To differentiate the product of two functions, we use the product rule:
Let u = e^-2x and v = Sec(3x).
Then, dy/dx = u(dv/dx) + v(du/dx).
Now, let's find du/dx and dv/dx:
To find du/dx, we differentiate u = e^-2x using the chain rule:
du/dx = (-2)e^-2x = -2e^-2x.
To find dv/dx, we differentiate v = Sec(3x) using the chain rule:
dv/dx = Sec(3x) * Tan(3x) * 3 = 3Sec(3x)Tan(3x).
Substituting the values into the product rule formula, we have:
dy/dx = (e^-2x)(3Sec(3x)Tan(3x)) + (Sec(3x))(-2e^-2x).
Simplifying the expression, we get:
dy/dx = 3e^-2xSec(3x)Tan(3x) - 2e^-2xSec(3x).
Therefore, the correct answer is A. 3e^-2xSec(3x)Tan(3x) - 2e^-2xSec(3x).
2.) To find dy/dx, we can use the chain rule applied to the natural logarithmic function.
The given function is y = Ln(2x^3 + 3x).
To differentiate y with respect to x, we have:
dy/dx = (1 / (2x^3 + 3x)) * (d(2x^3 + 3x) / dx).
Let's find d(2x^3 + 3x) / dx:
d(2x^3 + 3x) / dx = 6x^2 + 3.
Substituting this back into the original equation, we have:
dy/dx = (1 / (2x^3 + 3x)) * (6x^2 + 3).
Simplifying the expression, we get:
dy/dx = (6x^2 + 3) / (2x^3 + 3x).
Therefore, the correct answer is C. (6x^2 + 3) / (2x^3 + 3x).
3.) To find the angle that the tangent line makes with the radius vector, we need to find the derivative of the polar equation r = 3cos(theta) and evaluate it when theta = 120 degrees.
The polar equation is given as r = 3cos(theta).
To find the angle, we need to calculate dy/dx using the polar derivative formula:
dy/dx = (dy/dtheta) / (dx/dtheta).
First, let's find dy/dtheta by differentiating both sides of the polar equation with respect to theta:
dy/dtheta = 3(-sin(theta)).
Next, let's find dx/dtheta by differentiating both sides of the polar equation with respect to theta:
dx/dtheta = 3cos(theta).
Now, substitute these values into the formula for dy/dx:
dy/dx = (3(-sin(theta))) / (3cos(theta)).
Simplifying the expression, we have:
dy/dx = -tan(theta).
When theta = 120 degrees, we need to find tan(120 degrees).
Using a calculator or trigonometric identities, we find that tan(120 degrees) = -√3.
Therefore, the correct answer is D. 90 degrees.
4.) To calculate the angular acceleration when t = 2sec, we need to differentiate the given equation theta = 2^t - 6t with respect to time (t).
The given equation is theta = 2^t - 6t.
To find the angular acceleration, we need to find the second derivative, d^2(theta)/dt^2.
First, differentiate theta with respect to t to find the first derivative:
d(theta)/dt = 2^t * ln(2) - 6.
Next, differentiate d(theta)/dt with respect to t to find the second derivative:
d^2(theta)/dt^2 = (2^t * ln(2))' - (6)'.
Differentiating each term, we have:
d^2(theta)/dt^2 = (2^t)' * ln(2) + 2^t * ln(2)' - 0.
Now, let's find the derivatives of each term:
(2^t)' = (ln(2) * 2^t) * (1) = ln(2) * 2^t.
(ln(2))' = 0 since ln(2) is a constant.
Substituting these values back into the equation for the second derivative, we have:
d^2(theta)/dt^2 = ln(2) * 2^t * ln(2) + 0.
Simplifying the expression, we get:
d^2(theta)/dt^2 = ln(2)^2 * 2^t.
When t = 2sec, the expression becomes:
d^2(theta)/dt^2 = ln(2)^2 * 2^2 = ln(2)^2 * 4.
Therefore, the correct answer is B. 12 radians per sec^2.