the tangent to the curve y=a/x +bx at (1,3) is parallel to the line whose equation is y=2x+1. find the values of a and b.
First of all, the point (1,3) is on the curve, so
3 = a/1 + b(1)
a+b = 3
dy/dx = -a/x^2 + b
at (1,3) , dy/dx = -a + b
but it is parallel to y = 2x+1, so the slope is 2
-a + b = 2
add our two equations in a and b:
2b = 5
b = 5/2
sub back into a+b=3
a+5/2 = 3
a = 1/2
a = 1/2, b = 5/2
To find the values of a and b, we need to use two conditions:
1. The tangent to the curve y = a/x + bx at (1, 3) is parallel to the line y = 2x + 1.
2. The point (1, 3) lies on the curve.
Let's begin by finding the derivative of the curve y = a/x + bx, which will give us the slope of the tangent line at any given point on the curve.
1. Differentiating y = a/x + bx with respect to x:
dy/dx = -a/x^2 + b
The slope of the tangent line is the value of dy/dx at the point (1,3). So, we substitute x = 1 into dy/dx:
dy/dx = -a/1^2 + b = -a + b
Since the tangent line is parallel to y = 2x + 1, its slope is 2. So, we set -a + b = 2.
2. Next, we'll substitute the coordinates of the point (1, 3) into the equation of the curve y = a/x + bx:
3 = a/1 + b(1)
3 = a + b
Now, we have a system of two equations:
-a + b = 2
a + b = 3
We can solve this system of equations using any method, such as substitution or elimination.
Adding the two equations:
-a + a + b + b = 2 + 3
2b = 5
b = 5/2
Substituting the value of b into one of the equations, to find a:
a + (5/2) = 3
a = 3 - 5/2
a = 6/2 - 5/2
a = 1/2
Therefore, the values of a and b are a = 1/2 and b = 5/2.
To find the values of 'a' and 'b' for the curve y = a/x + bx, we need to use the given information about the tangent line and the equation of the line y = 2x + 1.
We know that the tangent to the curve is parallel to the line y = 2x + 1. This means that the slopes of the tangent line and the line y = 2x + 1 are equal.
Let's first find the slope of the line y = 2x + 1. We can see that the coefficient of 'x' is the slope, so the slope of the line y = 2x + 1 is 2.
Next, we need to find the slope of the tangent line to the curve y = a/x + bx. To do that, we differentiate the equation of the curve with respect to 'x'.
Differentiating y = a/x + bx:
dy/dx = -a/x^2 + b
Since the tangent line has the same slope as the line y = 2x + 1, we can set the derivative equal to 2 and solve for 'a' and 'b'.
- a/x^2 + b = 2
To simplify the equation, we need to eliminate 'x' from the equation. Since we are given that the tangent line passes through the point (1,3), we can substitute these values into the equation.
- a/(1)^2 + b = 2
-a + b = 2
We now have a system of two equations:
-a + b = 2 (Equation 1)
a/b = 3 (Equation 2)
To solve this system, we can use substitution or elimination. We will use substitution:
Rearrange Equation 2 to solve for 'a':
a = 3b
Substitute the value of 'a' in Equation 1:
-(3b) + b = 2
-2b = 2
b = -1
Substitute the value of 'b' into Equation 2:
a/(-1) = 3
a = -3
Therefore, the values of 'a' and 'b' are a = -3 and b = -1.