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
We need to find a and b, so we will look for two equations.
The first one is simply the point (1,1), which tells us that 1=y(1), or
1=a(1^3)+b(1)=a+b, or
a+b=1............(1)
The second one is the fact that the gradient (dy/dx) at (1,1)=-5.
Since dy/dx = 3ax²+b, we have
-5 = 3a(1^2)+b = 3a+b
or
3a+b = -5...........(2)
Solve for a and b given the system of equations:
a+b=1............(1)
3a+b = -5...........(2)
You can solve by elimination.
a=-3 ,b=4
Well, calculating the values of a and b might not be as simple as telling a joke, but let me give it a try!
To find the gradient of the tangent to the curve at the point (1, 1), we need to take the derivative of the curve equation with respect to x. If y = ax^3 + bx, then dy/dx = 3ax^2 + b.
Since we want the gradient to be -5 at x = 1, we can set up the equation as follows:
3a(1)^2 + b = -5
Simplifying, we have:
3a + b = -5
Now, we also know that the point (1, 1) lies on the curve, so we can substitute x = 1 and y = 1 in the original equation:
1 = a(1)^3 + b
1 = a + b
Now, we have a system of two equations:
3a + b = -5
a + b = 1
Let's solve this system. Subtracting the second equation from the first, we get:
2a = -6
Dividing both sides by 2, we find:
a = -3
Substituting this value back into the second equation, we can calculate b:
-3 + b = 1
b = 4
So, the values of a and b are -3 and 4, respectively. Now, that wasn't too funny, was it? But I hope it helps!
To find the values of a and b, we can use the gradient formula for a curve at a given point. The gradient (or slope) of a curve at a specific point can be found by taking the derivative of the equation of the curve and then evaluating it at that point.
Given that the curve is described by the equation y = ax^3 + bx, let's find the derivative of this equation with respect to x.
Differentiating y = ax^3 + bx:
dy/dx = 3ax^2 + b
Now, we know that the gradient at the point (1, 1) is -5. Let's substitute x = 1 and y = 1 into the derivative equation and solve for a and b.
-5 = 3a(1)^2 + b
-5 = 3a + b
We have a system of two equations:
1) -5 = 3a + b
2) 1 = a(1)^3 + b
From equation 1, we can express b in terms of a:
b = -5 - 3a
Substitute this expression for b into equation 2:
1 = a - 5 - 3a
Now solve for a:
1 = -2a - 5
2a = -6
a = -3
Once we have the value of a, we can substitute it back into equation 1 to find b:
-5 = 3(-3) + b
-5 = -9 + b
4 = b
Therefore, the values of a and b are a = -3 and b = 4, respectively.