18. Which is the polar form of the parametric equations x = 3t and y = t ^ 2 ?
To find the polar form of the parametric equations, we can use the following conversions:
$r=\sqrt{x^2+y^2}$ and $\tan \theta =\frac{y}{x}$
Substituting the given parametric equations:
$r=\sqrt{(3t)^2+(t^2)^2} = \sqrt{9t^2 + t^4} = t\sqrt{9 + t^2}$
$\tan \theta =\frac{t^2}{3t} = \frac{t}{3}$
Therefore, the polar form of the parametric equations is:
$$ \boxed{r = t\sqrt{9 + t^2}, \quad \theta = \arctan \frac{t}{3}} $$
To express the given parametric equations x = 3t and y = t^2 in polar form, we can substitute these equations into the polar coordinate conversions:
x = r*cos(θ)
y = r*sin(θ)
For convenience, let's solve for r and θ separately:
From x = 3t, we have:
r*cos(θ) = 3t
r = 3t / cos(θ) ----(1)
From y = t^2, we have:
r*sin(θ) = t^2
r = t^2 / sin(θ) ----(2)
Now, equating (1) and (2), we can find the relationship between t, r, and θ:
3t / cos(θ) = t^2 / sin(θ)
Cross-multiplying, we get:
3t * sin(θ) = t^2 * cos(θ)
Dividing both sides by t, we get:
3 * sin(θ) = t * cos(θ)
Now, since t represents the parameter and is typically expressed in terms of θ, we can solve for t in terms of θ:
t = (3 * sin(θ)) / cos(θ)
t = 3 * tan(θ) -----(3)
Finally, we substitute equation (3) back into either equation (1) or (2) to solve for r in terms of θ:
Using equation (1):
r = 3t / cos(θ)
r = 3(3 * tan(θ)) / cos(θ)
r = 9 * tan(θ) / cos(θ)
So, the polar form of the parametric equations x = 3t and y = t^2 is:
r = 9 * tan(θ) / cos(θ)
To find the polar form of the given parametric equations x = 3t and y = t^2, we need to express both x and y in terms of the polar coordinates r and θ.
We can start by eliminating the parameter t. From the first equation x = 3t, we can rearrange it as t = x/3. Now, substitute this value of t into the second equation:
y = (x/3)^2 = x^2/9
Next, we need to express x and y in terms of r and θ. In the polar coordinate system, x can be represented as rcosθ and y as rsinθ. Therefore:
x = rcosθ
y = rsinθ
Substituting these values into the above equation, we get:
rsinθ = (rcosθ)^2 / 9
Simplifying further:
rsinθ = r^2cos^2θ / 9
rsinθ = r^2cos^2θ / 9
9rsinθ = r^2cos^2θ
Now, we can use the Pythagorean identity sin^2θ + cos^2θ = 1 to eliminate the trigonometric functions:
9rsinθ = r^2(1 - sin^2θ)
9rsinθ = r^2 - r^2sin^2θ
9rsinθ = r^2(1 - sin^2θ)
9rsinθ = r^2(1 - (1 - cos^2θ))
9rsinθ = r^2cos^2θ
Finally, expressing r in terms of θ, we obtain the polar form:
r^2 = 9rsinθ / cos^2θ
r = 3√(sinθ / cos^2θ)
Hence, the polar form of the given parametric equations x = 3t and y = t^2 is r = 3√(sinθ / cos^2θ).