if the line 3x-4y=0 is tangent in the first quadrant to the curve y=x^3+k, find k.
all i know is that its related to finding a derivative and i m basically completely lost....can please someone provide me some help?????
the derivative gives the slope of the tangent line to the curve, at any point (x,y)
The line 3x-4y=0 has slope= 3/4
At any point on the curve (x,x^3 + k) the slope of the tangent line is 3x^2
So, you want to find x where 3x^2 = 3/4
x = 1/2
so, now you have y = 1/2^3 + k = 1/8 + k
What point on the line has y = 1/8 + k?
3* 1/2 - 4(1/8 + k) = 0
3/2 - 1/2 - 4k = 0
4k = 1
k = 1/4
So, y = x^3 + 1/4
At x = 1/2, y = 1/8 + 1/4 = 3/8
Note that at the point (1/2 , 3/8) 3x-4y = 0
Note on missing answer.
x can also be -1/2 if 3x^2 = 3/4
In that case, k = -1/4
and the point of tangency is (-1/2 , -3/8)
Note also, that if the line had been 3x-4y=1, the same line would have touched the curve at both points, with k=0.
thanks so much!
1/4
To find the value of k, you need to determine the derivative of the curve y=x^3+k and find the slope of the tangent line.
Here are the steps to follow:
1. Start by finding the derivative of the curve y=x^3+k. Since k is a constant, the derivative of k is simply 0. The derivative of x^3 is 3x^2. Therefore, the derivative of y=x^3+k is dy/dx = 3x^2.
2. The slope of the tangent line to the curve y=x^3+k is equal to the derivative dy/dx evaluated at the point of tangency. Since the line 3x-4y=0 is tangent to the curve, this means that the slope of the line is equal to the slope of the curve at that point.
3. Substitute the coordinates of the point of tangency into the equation 3x-4y=0. Since the line is in the first quadrant, both x and y should be positive. For convenience, let's assume that the point of tangency is (a, b). So, we have 3a - 4b = 0.
4. Substitute the expression 3x^2 for dy/dx in the equation 3a - 4b = 0. This gives 3a - 4b = 3a^2.
5. Solve the equation 3a - 4b = 3a^2 for b to get the value of k. Rearrange the equation as 3a^2 - 3a + 4b = 0 and then solve for b using quadratic formula or factoring.
Once you've calculated the value of k, you will have found the value that makes the line 3x-4y=0 tangent to the curve y=x^3+k in the first quadrant.