The gradient of the tangent to curve y = ax^3 + bx at the point (1,1) is -5. Calculate the values of a and b.
dy/dx = 3a^2 + b
-5 =
1 = a + b
-5 = 3a + b
solve the system for a and b
To find the values of a and b, we can start by taking the derivative of the equation y = ax^3 + bx with respect to x.
Taking the derivative, we get:
dy/dx = 3ax^2 + b
We are given that the gradient of the tangent at the point (1,1) is -5. So, at x = 1, the derivative should equal -5.
Substituting x = 1 in the derivative equation, we get:
-5 = 3a(1)^2 + b
-5 = 3a + b
We have two equations:
1) -5 = 3a + b
2) 1 = a(1)^3 + b
To solve these equations, we can multiply the first equation by -1 and add it to the second equation:
-5 = 3a + b (1)
1 = a + b (2)
-(-5) = -3a - b
5 = 3a + b
Adding equations (1) and (2), we get:
(3a + b) + (a + b) = -5 + 1
4a + 2b = -4
2a + b = -2 (3)
Now, we can solve equation (3) along with equation (2) using the method of substitution or elimination.
Substituting b = -2 - 2a from equation (3) into equation (2), we have:
1 = a + (-2 - 2a)
1 = a - 2 - 2a
1 = -a - 2
a = -3
Substituting the value of a back into equation (2), we get:
1 = -3 + b
b = 4
Therefore, a = -3 and b = 4.
To find the values of a and b, we can use the given information and equations relating to gradients and derivatives.
We are given that the gradient of the tangent to the curve is -5 at the point (1,1). The gradient or slope of a curve at a given point can be found by taking the derivative of the equation of the curve with respect to x.
The derivative of the curve y = ax^3 + bx with respect to x is given by:
dy/dx = 3ax^2 + b
We can substitute the values of x and y from the given point (1,1) into this equation:
-5 = 3a(1)^2 + b
-5 = 3a + b
Now we have a system of two equations:
-5 = 3a + b (Equation 1)
1 = a(1)^3 + b (Equation 2)
We can solve this system of equations to find the values of a and b. Here's one method:
1. Solve Equation 2 for b:
b = 1 - a
2. Substitute this expression for b in Equation 1:
-5 = 3a + (1 - a)
3. Simplify and solve for a:
-5 = 2a + 1
2a = -6
a = -3
4. Substitute the value of a back into Equation 2 to find b:
1 = (-3)(1)^3 + b
1 = -3 + b
b = 1 + 3
b = 4
Therefore, the values of a and b are:
a = -3
b = 4