The gradient of the tangent to the curve y= ax^3 + bx at the point ( 1, 1) is -5. Calculate the values of a and b
-4 and 3
To find the values of a and b, we need to use calculus.
First, we need to find the derivative of the curve. The derivative gives us the gradient of the tangent at any point on the curve.
Given that y = ax^3 + bx, we can find the derivative dy/dx by differentiating each term with respect to x:
dy/dx = 3ax^2 + b
Now we have the derivative equation.
Next, we need to find the gradient at the point (1, 1). To do this, we substitute x = 1 into the derivative equation:
dy/dx = 3a(1)^2 + b
dy/dx = 3a + b
We are told that the gradient of the tangent at (1, 1) is -5. So we can write this equation as:
-5 = 3a + b
This gives us one equation involving a and b.
To solve for a and b, we need another equation. We can use the fact that (1, 1) lies on the curve y = ax^3 + bx. Substituting x = 1 and y = 1 into this equation, we have:
1 = a(1)^3 + b
1 = a + b
This gives us a second equation involving a and b.
Now we have a system of two equations:
-5 = 3a + b
1 = a + b
We can solve this system to find the values of a and b.
One way to solve this is by substitution. Rearrange the second equation to get b in terms of a:
b = 1 - a
Substitute this expression for b in the first equation:
-5 = 3a + (1 - a)
Simplify the equation:
-5 = 2a + 1
Subtract 1 from both sides:
-6 = 2a
Divide both sides by 2:
-3 = a
Substitute this value back into the second equation to find b:
1 = -3 + b
Rearrange the equation to get b:
b = 4
Therefore, the values of a and b are a = -3 and b = 4, respectively.