If the line 3x-4y = 0 is tangent in the first quadrant to the curve y = x^3 + k, then k is
4 y = 3 x + 0
y = (3/4) x
slope = m = 3/4
y = x^3 + k
dy/dx = slope = 3 x^2
so
3 x^2 = 3/4
x^2 = 1/4
x = 1/2 (-1/2 is not in first quadrant)
then
y = (1/2)^3 + k
y = 1/8 + k and y = 3x/4
so
3 x/4 = 1/8 + k
but x = 1/2
3/8 = 1/8 + k
k = 2/8 = 1/4
Well, isn't that a tangential question! Let's find out what value of k makes that line a perfect tangent to the curve.
To determine if the line 3x - 4y = 0 is a tangent to y = x^3 + k, we need to find the point of tangency. In other words, we need to find the x-coordinate where the line and the curve intersect. Since the line is in the first quadrant, we can assume that x is positive.
Substituting the equation of the line into the equation of the curve, we get 3x - 4(x^3 + k) = 0. Simplifying this equation, we have:
3x - 4x^3 - 4k = 0
Now, let's solve for x. Unfortunately, solving this equation is a bit complex and not as straightforward as some others. But hey, if math was always straightforward, it wouldn't be as much fun, right?
So, instead of finding an exact solution for x, let's approximate it using some numerical methods, such as Newton's method. We can use computational tools to find the approximate value of x.
Once we find the approximate value of x, we can substitute it back into either the equation of the line or the curve to find the corresponding y-coordinate. From there, we can determine the value of k.
But hey, don't stress too much! Remember, math is just a little funny sometimes. Now, if only I could come up with a joke about tangents...
To find the value of k, we need to determine the condition for the line 3x - 4y = 0 to be tangent to the curve y = x^3 + k in the first quadrant.
Step 1: Find the derivative of the curve
Differentiating y = x^3 + k with respect to x, we get:
dy/dx = 3x^2
Step 2: Determine the slope of the tangent line
The slope of the tangent line is equal to the derivative of the curve evaluated at the point of tangency. In this case, the tangent line passes through a point in the first quadrant.
Let's find the point of tangency by solving the system of equations formed by the line and the curve:
3x - 4y = 0 ...(1)
y = x^3 + k ...(2)
Substituting equation (2) into equation (1), we get:
3x - 4(x^3 + k) = 0
3x - 4x^3 - 4k = 0
Step 3: Solve for the x-coordinate of the point of tangency
Rearranging the equation, we get:
4x^3 - 3x + 4k = 0 ...(3)
Since this is a cubic equation, finding its roots analytically is challenging. We can use numerical methods such as Newton's method or graphing to approximate the solutions.
Once we find the x-coordinate of the point of tangency, let's call it x_t, we can substitute it back into equation (2) to find the corresponding y-coordinate.
Step 4: Evaluate the derivative at the point of tangency
Substitute the x-coordinate of the point of tangency (x_t) into the derivative dy/dx = 3x^2, and calculate the derivative at that point: dy/dx = 3(x_t)^2.
Step 5: Set the slope of the tangent line equal to the derivative
Since the line 3x - 4y = 0 is tangent to the curve y = x^3 + k, the slope of the tangent line must be equal to the derivative at the point of tangency. Therefore, we need to solve the equation:
3(x_t)^2 = 3x - 4y
Step 6: Substitute the coordinate (x_t, y_t) of the point of tangency
Substitute the x-coordinate (x_t) and y-coordinate (y_t = (x_t)^3 + k) of the point of tangency into the equation 3(x_t)^2 = 3x - 4y, and solve for k.
k = y_t - (3/4) * x_t
Note: Steps 3 and 6 involve finding the approximate solution to a cubic equation, which can be computationally intensive.
To find the value of k when the line 3x-4y=0 is tangent to the curve y=x^3+k in the first quadrant, we can use the concept of tangency and slope of the curve.
When a line is tangent to a curve, it means that they touch at a single point and have the same slope at that point. Therefore, in order for the line 3x-4y=0 to be tangent to the curve y=x^3+k, we need to find a point that satisfies both the equation of the line and the equation of the curve, and also has the same slope.
Step 1: Set up the equations:
We have the equation of the line: 3x - 4y = 0
And the equation of the curve: y = x^3 + k
Step 2: Find the point of tangency:
Substitute the equation of the line into the equation of the curve:
3x - 4y = 0
4y = 3x
y = (3/4)x
Now substitute this expression for y into the equation of the curve:
(3/4)x = x^3 + k
Step 3: Solve for x:
Rearrange the equation to have zero on one side:
0 = x^3 - (3/4)x + k
Step 4: Evaluate the slope at the point of tangency:
To determine the slope of the curve at the point of tangency, we need to find the derivative of the curve function.
The derivative of y = x^3 + k is given by:
dy/dx = 3x^2
Substitute the x-value of the point of tangency into the derivative to find the slope:
Slope = 3(x-value of tangency)^2
Step 5: Equate slopes and solve for k:
Set the slope of the line (3/4) equal to the slope of the curve (3(x-value of tangency)^2). This will give us the x-value of the point of tangency:
3/4 = 3(x-value of tangency)^2
Solve for the x-value of the point of tangency:
(x-value of tangency)^2 = 1/4
x-value of tangency = 1/2 or -1/2
Step 6: Substitute the x-value back into the equation of the curve to find k:
Substitute x = 1/2 into y = x^3 + k:
y = (1/2)^3 + k
y = 1/8 + k
Substitute x = -1/2 into y = x^3 + k:
y = (-1/2)^3 + k
y = -1/8 + k
Since the line is tangent in the first quadrant, the y-value of the point of tangency must be positive. Therefore, we take the solution where x = 1/2:
y = 1/8 + k
Since y > 0, we can conclude that k > -1/8.
Therefore, k is any value greater than -1/8.