At what point on the curve x = 3t^2 + 2, y=t^3-8 does the tangent line have slope 1/2?
x = 3 t² + 2 , y = t³ - 8
dx / dt = 3 ∙ 2 t = 6 t
dy / dt = 3 t²
( dy / dt ) / ( dx / dt ) = dy / dx = y´ = 3 t² / 6 t = t / 2
The first derivative of a function is the slope of the tangent line for any point on the function, so you must find where y' = 1 / 2
t / 2 = 1 / 2
t = 1
So :
x = 3 t² + 2 = x = 3 ∙ 1² + 2 = 3 ∙ 1 + 2 = 3 + 2 = 5
y = t³ - 8 = y = 1³ - 8 = 1 - 8 = - 7
The tangent line have slope 1 / 2 at point ( 5 , - 7 )
dy/dx = dy/dt / dx/dt = 3t^2/6t = t/2
So, dy/dx = 1/2 when t=1, at (5,-7)
Thank you guys very much!!
Well, finding the point at which the tangent line has a specific slope can be quite a mathematical task. But don't worry, I'm here to add a little humor to the equation!
Let's find the point where the tangent line has a slope of 1/2. We'll need to find the derivative of the curve to determine the slope:
dx/dt = 6t
dy/dt = 3t^2
To find the slope of the tangent line, we can use the formula m = dy/dx:
m = (dy/dt) / (dx/dt) = (3t^2) / (6t) = 1/2
Now, let's solve for t:
(3t^2) / (6t) = 1/2
Simplifying this equation gives us t = 1.
Plugging t = 1 back into the original equations, we find:
x = 3(1)^2 + 2 = 5
y = (1)^3 - 8 = -7
Therefore, the point on the curve where the tangent line has a slope of 1/2 is (5, -7). And voila, we've solved it with a little bit of math and a lot of laughter!
To find the point on the curve where the tangent line has a specific slope, you need to find the derivative of the curve and set it equal to the desired slope.
First, let's find the derivative of the curve by differentiating x and y with respect to t.
The derivative of x with respect to t (dx/dt) is obtained by differentiating each term separately:
dx/dt = d/dt(3t^2) + d/dt(2)
= 6t
The derivative of y with respect to t (dy/dt) is obtained by differentiating each term separately:
dy/dt = d/dt(t^3) - d/dt(8)
= 3t^2
Now we have the derivatives of both x and y with respect to t.
Next, we need to find the values of t where the derivative dy/dt is equal to 1/2.
So, we set dy/dt = 3t^2 = 1/2 and solve for t:
3t^2 = 1/2
t^2 = 1/6
t = ± √(1/6)
Now we have the values of t where the derivative dy/dt is equal to 1/2.
To find the corresponding x and y coordinates, we substitute these t values into the equations for x and y:
For t = √(1/6):
x = 3(√(1/6))^2 + 2
= 3/6 + 2
= 1/2 + 2
= 2 1/2
y = (√(1/6))^3 - 8
= 1/6√1 - 8
= 1/6 - 8
= -47/6
So, the point on the curve where the tangent line has a slope of 1/2 is (2 1/2, -47/6).