The line y=2x + b is tangent to the graph y= sqrt x at the point P= (a,sqrt a). Find P and determine b
Recall that sqrt(x) = x^(1/2), so the slope of the graph at any point x is 1/(2*sqrt(x)).
So, at point P, x=a, slope is 1/(2*sqrt(a)).
So, the line is y=2x + b, meaning that the slope is 2. That means that 2 = 1/(2*sqrt(a)). So, a=1/16.
P = (1/16, 1/4)
y=2x+b
1/4 = 2(1/16)+b
1/4 = 1/8 + b
1/8 = b
y = 2x + 1/8
Well, finding P is quite simple. Since the line y=2x+b is tangent to the graph y=sqrt(x) at point P=(a,sqrt(a)), it means that the slopes of both the line and the curve at that point are equal.
The slope of the line y=2x+b is 2.
To find the slope of the curve y=sqrt(x) at point P=(a,sqrt(a)), we need to take the derivative of y=sqrt(x) and plug in a for x:
(dy/dx) = (1/2) * (x^(-1/2))
(dy/dx) = 1/(2*sqrt(x))
(dy/dx) = 1/(2*sqrt(a))
Now, we can equate the slopes:
2 = 1/(2*sqrt(a))
4*sqrt(a) = 1
sqrt(a) = 1/4
a = 1/16
So, P = (1/16, 1/4)
Now, to determine the value of b, we can substitute the x and y values of P into the equation of the line:
1/4 = 2 * (1/16) + b
1/4 = 1/8 + b
b = 1/4 - 1/8
b = 2/8 - 1/8
b = 1/8
Therefore, P = (1/16, 1/4) and b = 1/8.
To find the point P where the line y=2x+b is tangent to the graph y=sqrt(x), we need to set the two equations equal to each other and solve for x and y.
1. Start with the equation for the line: y = 2x + b
2. Substitute y=sqrt(x) into the equation for the line: sqrt(x) = 2x + b
3. Square both sides of the equation to eliminate the square root: x = (2x + b)^2
4. Expand the right side of the equation: x = 4x^2 + 4bx + b^2
5. Rewrite the equation in standard form: 4x^2 + (4b-1)x + b^2 - x = 0
6. Since the line is tangent to the graph, it will only touch the graph at one point. This means the quadratic equation has a double root, so the discriminant is zero.
7. Set the discriminant to zero: (4b-1)^2 - 4(4)(b^2 - x) = 0
8. Simplify the equation: (4b-1)^2 - 16b^2 + 16x = 0
9. Expand and rearrange the equation: 16b^2 - 8b + 1 - 16b^2 + 16x = 0
10. Combine like terms: 16x - 8b + 1 = 0
11. To solve for x, we can isolate it by moving the other terms to the other side of the equation: 16x = 8b - 1
12. Divide both sides by 16 to solve for x: x = (8b - 1)/16
13. Since P = (a, sqrt(a)), substitute this into the equation for x: sqrt(a) = (8b - 1)/16
14. Square both sides of the equation to eliminate the square root: a = (8b - 1)^2/256
15. Simplify the equation: a = (64b^2 - 16b + 1)/256
So, point P is (a, sqrt(a)) = ((64b^2 - 16b + 1)/256, sqrt((64b^2 - 16b + 1)/256)).
To find the point of tangency P = (a, √a), we need to find the value of a. We know that the line y = 2x + b is tangent to the graph y = √x, so the slope of the line 2 (m1) must be equal to the slope of the curve at the point P (m2).
The slope of the line 2x + b can be determined by looking at its coefficient of x, which is 2.
To find the slope of the curve y = √x at the point P = (a, √a), we take the derivative of y = √x with respect to x:
dy/dx = (1/2) * x^(-1/2)
Now, we substitute x = a into the derivative to get the slope at P:
m2 = (1/2) * a^(-1/2)
Since m1 = m2, we have:
2 = (1/2) * a^(-1/2)
To solve for a, we can multiply both sides of the equation by 2 to get:
4 = a^(-1/2)
We can then square both sides of the equation to eliminate the square root:
16 = a^(-1)
Taking the reciprocal of both sides, we get:
1/16 = a
Therefore, the point of tangency P is (a, √a) = (1/16, √(1/16)) = (1/16, 1/4).
To determine the value of b, we substitute the x-coordinate of P into the equation of the line y = 2x + b:
b = y - 2x = (√a) - 2a = (√(1/16)) - (2 * (1/16)) = (1/4) - (1/8) = 1/8
Thus, the value of b is 1/8.