If the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k, then k is
a: 1/2
b: 1/4
c: 0
d: -1/8
e: -1/2
if y = x^3 + k
dy/dx = 3x^2 , which of course is the slope of the tangent
but the slope of the given line is 3/4 , so ...
3x^2 = 3/4
x^2 = 1/4
x = ± 1/2 , but since we are restricted to the first quad,
x = 1/2
if x=1/2, 3(1/2) - 4y = 0
y = 3/8
The point of contact of the tangent is (1/2 , 3/8)
so in the cubic ...
3/8 = (1/8) + k
k = 3/8-1/8 = 1/4
Thank youu(:
To find the value of k, we need to determine the point(s) where the line and the curve intersect in the first quadrant.
First, let's solve the system of equations formed by the line and the curve:
Line equation: 3x - 4y = 0
Curve equation: y = x^3 + k
Substituting y in the line equation with the expression from the curve equation:
3x - 4(x^3 + k) = 0
Expanding and rearranging the equation:
3x - 4x^3 - 4k = 0
Next, let's find the derivative of the curve equation:
y = x^3 + k
dy/dx = 3x^2
Since the line is tangent to the curve, the derivative of the curve equation at the intersection point will be equal to the slope of the line. The slope of the line can be determined by rearranging the line equation in slope-intercept form:
3x - 4y = 0
4y = 3x
y = (3/4)x
Comparing the slope of the line (3/4) to the derivative of the curve equation (3x^2), we can set them equal to each other:
3x^2 = 3/4
Simplifying the equation:
x^2 = 1/4
Taking the square root of both sides:
x = ±1/2
Since we are only interested in the intersection point(s) in the first quadrant, we take x = 1/2.
Now, substituting x = 1/2 back into the equation 3x - 4x^3 - 4k = 0:
3(1/2) - 4(1/2)^3 - 4k = 0
3/2 - 1/8 - 4k = 0
3 - 1/4 - 32k = 0
12 - 1 - 32k = 0
-1 - 32k = 0
-32k = 1
k = -1/32
So, the value of k is -1/32.
Therefore, the correct option is d: -1/8.
To determine the value of k, we need to find the point of tangency between the line and the curve.
Step 1: Finding the slope of the line
Since the line is given by the equation 3x - 4y = 0, we can rearrange it to find the slope-intercept form: y = (3/4)x.
Comparing this equation to y = mx + b, we can see that the slope (m) of the line is 3/4.
Step 2: Finding the derivative of the curve
The given curve is y = x^3 + k. To find the derivative, we differentiate the equation with respect to x.
dy/dx = 3x^2.
Step 3: Finding the x-coordinate of the point of tangency
To find the x-coordinate of the point of tangency, we set the slope of the line equal to the derivative of the curve:
3/4 = 3x^2.
Solving this equation, we get:
x^2 = 1/4
x = ±√(1/4)
x = ±1/2.
Since we're looking for a point in the first quadrant, we take x = 1/2.
Step 4: Finding the y-coordinate of the point of tangency
To find the y-coordinate of the point of tangency, we substitute the value of x into the equation for the curve:
y = (1/2)^3 + k
y = 1/8 + k
Step 5: Determining the value of k
Since the line is tangent to the curve, the y-coordinate of the point of tangency should be equal to the y-intercept of the line (when x = 0).
Substituting x = 0, the equation of the line gives us:
-4y = 0
y = 0.
Comparing this to the y-coordinate of the point of tangency, 1/8 + k, we get:
0 = 1/8 + k.
Solving for k, we find:
k = -1/8.
Therefore, the answer is d: -1/8.