2- given the curve is described by the equation r=3cos ¥è, find the angle that the tangent line makes with the radius vector when ¥è=120¨¬.
A. 30¨¬
B. 45¨¬
C. 60¨¬
D. 90¨¬
not sure A or D
2.) 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)
3.)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
Thank you!
2 and 3 are right.
the first one is a trick question. It is a circle, so the tangent line is always 90 degrees to the radius
1) 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(θ) with respect to θ.
First, differentiate both sides with respect to θ:
d(r)/d(θ) = d(3cos(θ))/d(θ)
Since the derivative of cos(θ) with respect to θ is -sin(θ), we get:
d(r)/d(θ) = -3sin(θ)
Next, we can calculate the value of dy/dx (slope of the tangent line) at θ = 120° by applying some trigonometry. For a given polar point (r, θ), the corresponding Cartesian coordinates are (x, y), where x = rcos(θ) and y = rsin(θ). Since we are interested in the tangent line at θ = 120°, we can substitute θ = 120° into the equation:
x = 3cos(120°) = -1.5
y = 3sin(120°) = 2.598
So the point on the curve where θ = 120° is (-1.5, 2.598).
To find the slope of the tangent line (dy/dx), we can calculate the ratio of the change in y to the change in x as we approach the point on the curve. We can do this by converting the polar coordinates to Cartesian coordinates and then taking the derivative:
dy/dx = (dy/dθ)/(dx/dθ)
= (d(rsin(θ))/dθ) / (d(rcos(θ))/dθ)
= (d(3sin(θ))/dθ) / (d(3cos(θ))/dθ)
= (-3cos(θ)) / (-3sin(θ))
= cos(θ) / sin(θ)
= cot(θ)
Now, let's substitute θ = 120° to find the value of dy/dx:
dy/dx = cot(120°) = cos(120°) / sin(120°)
= (-0.5) / (sin(60°))
= (-0.5) / (sqrt(3) / 2)
= -1 / sqrt(3)
= -sqrt(3) / 3
The tangent line makes an angle θ' with the positive x-axis, where tan(θ') = dy/dx. So we need to find the angle whose tangent is -sqrt(3)/3:
θ' = atan(-sqrt(3)/3)
= -30°
Since we are looking for the acute angle, θ' = 30°.
Therefore, the correct answer is A. 30°.
2) To find dy/dx for y = e^(-2x) sec(3x), we have a product of two functions, e^(-2x) and sec(3x). We can use the product rule of differentiation.
dy/dx = (d(e^(-2x))/dx) * sec(3x) + e^(-2x) * (d(sec(3x))/dx)
The derivative of e^(-2x) with respect to x is -2e^(-2x) (by the chain rule), and the derivative of sec(3x) with respect to x is 3sec(3x)tan(3x) (by the chain rule).
Substituting these values into the equation, we get:
dy/dx = -2e^(-2x) * sec(3x) + e^(-2x) * 3sec(3x)tan(3x)
= e^(-2x) * (-2sec(3x) + 3sec(3x)tan(3x))
Therefore, the correct answer is B. 3e^(-2x) sec(3x) tan(3x) - 2xe^(-2x) sec(3x).
3) To find dy/dx for y = ln(2x^3 + 3x), we can use the chain rule of differentiation.
dy/dx = d(ln(2x^3 + 3x))/dx
The derivative of ln(u) with respect to x is du/dx / u, where u represents the expression inside the ln function.
In this case, u = 2x^3 + 3x. Let's find du/dx:
du/dx = d(2x^3 + 3x)/dx
= 6x^2 + 3
Now, substitute back into the equation:
dy/dx = (6x^2 + 3)/(2x^3 + 3x)
Therefore, the correct answer is C. (2x^3 + 3x)/(6x^2 + 3).