A curve passes through the point (1,-11) and it's gradient at any point is ax^2 + b, where a and b are constants. The tangent to the curve at the point (2,-16) is parallel to the x-axis. Find
i) the values of a and b
ii) the equation of the curve
since dy/dx = ax^2 + b
the equation must have been
y = (1/3)a x^3 + bx + c
at (1,-11) ---> -11 = (1/3) a + b + c
or
a + 3b + 3c = -33 (#1)
at (2, -16) ---> -16 = (8/3)a + 2b + c
or
8a + 6b + 3c = -48 (#2)
#2 - #1 :
7a + 3b = -15 (#3)
We also know that at (2,-16), the slope is zero
ax^2 + b = 0
4a + b = 0
b = -4a
sub into #3
7a - 12a = -15
-5a=-15
a=3
then b= -12
and in #1
3 -36+ 3c = -33
3c = 0
c = 0
function is y = (1/3)(3)x^3 - 12x
y = x^3 - 12x
I like the way you solve it
i) To find the values of a and b, we can use the fact that the gradient of a curve at a given point is equal to the derivative of the equation of the curve at that point.
Given that the gradient of the curve at any point is ax^2 + b, we can set up the following equation:
dy/dx = ax^2 + b
By differentiating both sides of the equation, we get:
d^2y/dx^2 = 2ax
Since the tangent to the curve at the point (2,-16) is parallel to the x-axis, the slope of the tangent is 0. In other words, the gradient at that point is 0.
So, we have:
0 = a(2)^2 + b
0 = 4a + b
We also know that the curve passes through the point (1,-11). Plugging these coordinates into the equation of the curve, we get:
-11 = a(1)^2 + b
-11 = a + b
Now, we have a system of equations:
4a + b = 0
a + b = -11
To solve this system, we can use substitution or elimination method. I'll use the elimination method here:
Multiply the second equation by -4 to make the coefficients of 'b' equal:
-4(a + b) = -4(-11)
-4a - 4b = 44
Now, we can add this equation to the first equation:
(4a + b) + (-4a - 4b) = 0 + 44
-3b = 44
Divide both sides by -3:
b = -44/3
Substitute the value of b into one of the original equations to solve for a:
a + b = -11
a + (-44/3) = -11
a = -11 + 44/3
a = -11/3
Therefore, the values of a and b are a = -11/3 and b = -44/3.
ii) Now that we have the values of a and b, we can find the equation of the curve.
The equation of the curve can be written in the form:
y = f(x) = mx + c
where m is the gradient at any point and c is the y-intercept.
The gradient at any point is given as ax^2 + b, so the equation becomes:
y = (ax^2 + b)x + c
Plugging in the values of a = -11/3 and b = -44/3, we have:
y = (-11/3)x^3 + (-44/3)x + c
To find c, we can use the fact that the curve passes through the point (1,-11). Plugging in these coordinates, we get:
-11 = (-11/3)(1)^3 + (-44/3)(1) + c
-11 = -11/3 - 44/3 + c
-11 = -55/3 + c
To simplify, we need to find a common denominator:
-11 = (-55 + 3c)/3
Multiply both sides by 3:
-33 = -55 + 3c
Add 55 to both sides:
22 = 3c
Divide both sides by 3:
c = 22/3
Therefore, the equation of the curve is:
y = (-11/3)x^3 + (-44/3)x + 22/3
To find the values of a and b, we can use the fact that the gradient of the curve at any point is given by ax^2 + b.
(i) Since the tangent to the curve at the point (2,-16) is parallel to the x-axis, the gradient at that point must be 0. Therefore, we can set ax^2 + b equal to 0:
a(2)^2 + b = 0
4a + b = 0
We also know that the curve passes through the point (1,-11). Plugging in these coordinates into the equation of the curve, we have:
a(1)^2 + b = -11
a + b = -11
Now we can solve this system of equations to find the values of a and b. Subtracting the second equation from the first, we have:
4a + b - (a + b) = 0 - (-11)
3a = 11
a = 11/3
Plugging this value back into the second equation, we can solve for b:
(11/3) + b = -11
b = -11 - (11/3)
b = -33/3 - 11/3
b = -44/3
Therefore, the values of a and b are a = 11/3 and b = -44/3.
(ii) Now that we have the values of a and b, we can find the equation of the curve. The gradient of the curve at any point is ax^2 + b, so the equation of the curve can be obtained by integrating this expression with respect to x:
∫(ax^2 + b) dx = ∫(11/3)x^2 - (44/3) dx
= (11/3) * (x^3/3) - (44/3) * x + C
Where C is the constant of integration. Since the curve passes through the point (1,-11), we can substitute these coordinates into the equation to find the value of C:
(11/3) * (1^3/3) - (44/3) * 1 + C = -11
11/3 - 44/3 + C = -11
-33/3 + C = -11
C = -11 + 11 = 0
Therefore, the equation of the curve is:
(11/3) * (x^3/3) - (44/3) * x + 0
Thus, the equation of the curve is given by:
y = (11/3) * (x^3/3) - (44/3) * x