Find the cartesian equation of r = 8 sin thet + 8 cos theta

1 answer:

4 0

Use following substitutions:

r cosθ = x 
r sinθ = u 
r² = x² + y² 

r = 8 cosθ + 4 sinθ 

Multiply both sides by r 

r² = 8 r cosθ + 4 r sinθ 
x² + y² = 8x + 4y 

This is equation of circle. We can put in standard form: 

x² − 8x + y² − 4y = 0 
x² − 8x + 16 + y² − 4y + 4 = 16 + 4 
(x − 4)² + (y − 2)² = 20 

Circle centered at point (4, 2) with radius = √20

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Find the 2nd Derivative: f(x) = 3x⁴ + 2x² - 8x + 4

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Answer:

$f''(x)=36x^2+4$

Step-by-step explanation:

Let's start by finding the first derivative of $f(x)= 3x^4+2x^2-8x+4$ . We can do so by using the power rule for derivatives.

The power rule states that:

$\frac{d}{dx} (x^n) = n \times x^n^-^1$

This means that if you are taking the derivative of a function with powers, you can bring the power down and multiply it with the coefficient, then reduce the power by 1.

Another rule that we need to note is that the derivative of a constant is 0.

Let's apply the power rule to the function f(x).

$\frac{d}{dx} (3x^4+2x^2-8x+4)$

Bring the exponent down and multiply it with the coefficient. Then, reduce the power by 1.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = ((4)3x^4^-^1+(2)2x^2^-^1-(1)8x^1^-^1+(0)4)$

Simplify the equation.

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x^1-8x^0+0)$
$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8(1)+0)$

$\frac{d}{dx} (3x^4+2x^2-8x+4) = (12x^3+4x-8)$
$f'(x)=12x^3+4x-8$

Now, this is only the first derivative of the function f(x). Let's find the second derivative by applying the power rule once again, but this time to the first derivative, f'(x).