At what point on the curve
x = 3t2 + 7,y = t3 − 8
does the tangent line have slope 1/2
dy/dt = 3t^2
dx/dt = 6t
(dy/dt) / (dx/dt) = dy/dx = 3t^2/6t = t/2
so t/2 = 1/2 ===> t = 1
then:
x = 3(1)^2 + 7 = 10
y = 1^3 - 8 = -7
your point is (10,-7)
Well, finding the point on the curve where the tangent line has a specific slope can be quite tricky. But I'm here to make things fun, so let's give it a shot!
Let's find the slope of the tangent line first. To do that, we need to differentiate both x and y with respect to t:
dx/dt = 6t
dy/dt = 3t^2
Now, we want to find the value of t where the tangent line has a slope of 1/2. So, let's set dy/dt equal to 1/2:
1/2 = 3t^2
To solve this equation, let's multiply both sides by 2:
1 = 6t^2
Divide both sides by 6:
1/6 = t^2
Now, let's take the square root of both sides to find t:
√(1/6) = t
But wait a second! This equation has both a positive and a negative root, which means we might have two points on the curve where the tangent line has a slope of 1/2.
Calculating the square root of 1/6 gives us approximately ±0.408. So we have two values for t: -0.408 and 0.408.
Now, let's substitute these values of t back into the equations for x and y to find the corresponding points on the curve:
For t = -0.408:
x = 3*(-0.408)^2 + 7 ≈ 6.323
y = (-0.408)^3 - 8 ≈ -8.173
For t = 0.408:
x = 3*(0.408)^2 + 7 ≈ 7.677
y = (0.408)^3 - 8 ≈ -7.827
So it seems like the tangent line has a slope of 1/2 at the points (6.323, -8.173) and (7.677, -7.827) on the curve.
But hey, don't trust a clown when it comes to math! Double-check my calculations, just to be sure!
To find the point on the curve where the tangent line has a slope of 1/2, we need to find a value of t that satisfies this condition. Here's how we can do it step-by-step:
Step 1: Find the derivative of the curve.
The curve is given by the parametric equations x = 3t^2 + 7 and y = t^3 - 8. To find the derivative, we can differentiate these equations with respect to t:
dx/dt = 6t
dy/dt = 3t^2
Step 2: Write the equation of the tangent line.
The tangent line to the curve at any point is given by the equation (y - y0) = m(x - x0), where m is the slope of the tangent line and (x0, y0) is the point on the curve.
Step 3: Substitute the values into the equation.
We are given that the slope of the tangent line is 1/2. So, we can replace m with 1/2 in the equation from Step 2:
(y - y0) = (1/2)(x - x0)
Step 4: Substitute the expressions for x and y from the original parametric equations.
Using the expressions for x and y from the original parametric equations, we can substitute these into the equation from Step 3:
(t^3 - 8 - y0) = (1/2)(3t^2 + 7 - x0)
Step 5: Substitute y0 and x0 with the corresponding values of y and x for the point on the curve.
We don't know the values of y0 and x0 yet, but they represent the y and x values at a specific point on the curve. To find this point, we need to solve the following system of equations:
t^3 - 8 = y0
3t^2 + 7 = x0
Step 6: Solve the system of equations to find t.
By substituting the expression for y0 from Step 5 into the equation from Step 4, we get:
(t^3 - 8 - (t^3 - 8)) = (1/2)(3t^2 + 7 - x0)
0 = (1/2)(3t^2 + 7 - x0)
Simplifying this equation, we have:
3t^2 + 7 - x0 = 0
Step 7: Solve the equation from Step 6 for t.
Now, we can solve this equation for t:
3t^2 + 7 = x0
3t^2 = x0 - 7
t^2 = (x0 - 7)/3
t = sqrt((x0 - 7)/3)
Step 8: Plug in the value of t into the equations for x and y to find the coordinates of the point on the curve.
Using the expression for t from Step 7, we can now find the x and y coordinates of the point on the curve where the tangent line has a slope of 1/2:
x = 3t^2 + 7
x = 3((x0 - 7)/3) + 7
x = x0 - 7 + 7
x = x0
y = t^3 - 8
y = ((x0 - 7)/3)^3 - 8
So, the point on the curve where the tangent line has a slope of 1/2 is (x0, ((x0 - 7)/3)^3 - 8).
To find the point on the curve where the tangent line has a slope of 1/2, we need to differentiate the given equations and find the point where the derivative of y with respect to x is equal to 1/2.
Let's start by finding the derivatives of x and y with respect to t:
Given x = 3t^2 + 7, we can find dx/dt as follows:
dx/dt = d/dt(3t^2 + 7)
= 6t
Given y = t^3 - 8, we can find dy/dt as follows:
dy/dt = d/dt(t^3 - 8)
= 3t^2
Next, we need to find dy/dx by dividing dy/dt by dx/dt:
dy/dx = (dy/dt) / (dx/dt)
= (3t^2) / (6t)
= 1/2t
Now, we set dy/dx equal to 1/2 and solve for t:
1/2 = 1/2t
t = 1
Now that we have the value of t, we can substitute it back into the given equations to find the corresponding values of x and y:
For x = 3t^2 + 7:
x = 3(1^2) + 7
x = 10
For y = t^3 - 8:
y = 1^3 - 8
y = -7
Therefore, the point on the curve where the tangent line has a slope of 1/2 is (10, -7).