Show that the curve with parametric equations x=tcost, y=tsint, z=t lies on the cone z^2=x^2+y^2
Well, let's see if we can put a smile on that cone by showing that the given curve lies on it!
First, let's substitute the parametric equations into the equation of the cone:
z^2 = x^2 + y^2
(t)^2 = (tcos(t))^2 + (tsin(t))^2
Simplifying a bit:
t^2 = t^2 * cos^2(t) + t^2 * sin^2(t)
Now, since cos^2(t) + sin^2(t) = 1, we have:
t^2 = t^2 * (1)
or
t^2 = t^2
And guess what? That's true for any value of t!
So, the equation holds for all values of t, which means the curve with parametric equations x = tcos(t), y = tsin(t), z = t lies on the cone z^2 = x^2 + y^2.
Looks like the cone has a new companion to keep it company!
To show that the curve with parametric equations x = tcos(t), y = tsin(t), z = t lies on the cone z^2 = x^2 + y^2, we need to substitute the given equations into the equation of the cone and show that it holds true for any value of t.
Substituting the parametric equations into the equation of the cone:
(z^2) = (x^2) + (y^2)
Replacing x, y, and z with their respective parametric forms:
(t^2)^2 = (tcos(t))^2 + (tsin(t))^2
Simplifying the equation:
t^4 = t^2(cos^2(t) + sin^2(t))
Using the trigonometric identity cos^2(t) + sin^2(t) = 1:
t^4 = t^2(1)
t^4 = t^2
Dividing both sides of the equation by t^2:
t^2 = 1
Taking the square root of both sides:
t = ±1
Therefore, the curve with the given parametric equations lies on the cone z^2 = x^2 + y^2 for t = ±1.
To show that the curve with parametric equations x = tcos(t), y = tsin(t), and z = t lies on the cone z^2 = x^2 + y^2, we need to substitute these equations into the given relation and verify that the identity holds.
First, let's substitute x = tcos(t), y = tsin(t), and z = t into the equation z^2 = x^2 + y^2:
(t)^2 = (tcos(t))^2 + (tsin(t))^2
Simplifying the right side:
t^2 = t^2(cos(t))^2 + t^2(sin(t))^2
Using the trigonometric identity sin^2(t) + cos^2(t) = 1, we can simplify further:
t^2 = t^2[(cos(t))^2 + (sin(t))^2]
t^2 = t^2(1)
t^2 = t^2
The equation simplifies to a true statement, t^2 = t^2. Therefore, for any value of t, the curve with parametric equations x = tcos(t), y = tsin(t), and z = t lies on the cone z^2 = x^2 + y^2.
x^2 = t^2 cos^2
y^2 = t^2 sin^2
x^2 + y^2 = t^2(cos^2+sin^2) = t^2
but t^2 = z^2
so
x^2+y^2 = z^2