a) P is the point on the curve y= 2x^3 +kx -5 where x= 1 and gradient is -2.
Find
(i) The value of the constant k
(ii) The equation of the normal to the curve at P
y = 2x^3 + kx - 5
y' = 6x^2 + k
so, you want
6+k = -2
k = -8
so,
y = 2x^3-8x-5
since the slope is -2 at x=1,
y(1) = -11
The normal has slope m = 1/2, so at P(1,-11)
y+11 = 1/2 (x-1)
To find the value of the constant k and the equation of the normal to the curve at point P, we need to follow these steps:
(i) Finding the value of the constant k:
1. Start with the equation of the curve: y = 2x^3 + kx - 5.
2. Find the derivative of the equation with respect to x to get the gradient of the curve: dy/dx = 6x^2 + k.
3. Since we know that the gradient at point P is -2, we can substitute x = 1 into the derivative expression: dy/dx = 6(1)^2 + k = 6 + k.
4. Set the derived expression equal to the given gradient and solve for k: 6 + k = -2. Subtract 6 from both sides: k = -8.
The value of the constant k is -8.
(ii) Finding the equation of the normal to the curve at point P:
1. Using the value of k we found in step (i), substitute it back into the original equation of the curve: y = 2x^3 - 8x - 5.
2. Find the derivative of the curve equation: dy/dx = 6x^2 - 8.
3. To find the gradient of the normal, we need to take the negative reciprocal of the derivative at point P. So, the gradient of the normal (m) is -1 / (dy/dx).
At x = 1, dy/dx = 6(1)^2 - 8 = -2. Therefore, the gradient of the normal (m) is -1 / (-2) = 1/2.
The gradient of the normal is 1/2.
4. The equation of a straight line is y - y1 = m(x - x1), where (x1, y1) is a point on the line.
Since the line is the normal to the curve at point P(1, y), we can substitute the values into the equation: y - y1 = (1/2)(x - x1), where P(1, y).
Therefore, the equation of the normal to the curve at point P is y - y = (1/2)(x - 1).
Simplifying the equation, we get y = (1/2)x - 1/2 as the equation of the normal to the curve at point P.
To find the value of the constant k and the equation of the normal to the curve at point P, we need to use the given information - the point P on the curve and the gradient at that point.
Let's solve this step by step:
(i) Finding the value of the constant k:
We have the equation of the curve as y = 2x^3 + kx - 5. We are given that P is a point on the curve where x = 1 and the gradient is -2.
The gradient of a function at a particular point can be found by taking the derivative of the function with respect to x.
Differentiating the equation of the curve, y = 2x^3 + kx - 5, with respect to x, we get:
dy/dx = 6x^2 + k
We are given that the gradient at point P is -2. So, setting dy/dx = -2 and substituting x = 1, we have:
-2 = 6(1)^2 + k
-2 = 6 + k
To solve for k, subtract 6 from both sides:
-2 - 6 = k
-8 = k
Therefore, the value of the constant k is -8.
(ii) Finding the equation of the normal to the curve at point P:
To find the equation of the normal, we need to find the slope of the normal at point P. The slope of the normal is the negative reciprocal of the gradient.
Since the gradient at point P is -2, the slope of the normal is the negative reciprocal of -2, which is 1/2.
We also know that the normal passes through the point P, which has coordinates (1, y). To find y, substitute x = 1 into the equation of the curve and solve for y:
y = 2(1)^3 + (-8)(1) - 5
y = 2 - 8 - 5
y = -11
So, the coordinates of point P are (1, -11).
Using the slope-intercept form of the equation of a line, y = mx + c, where m is the slope and c is the y-intercept, we can write the equation of the normal:
y = (1/2)x + c
To find the value of c, substitute the coordinates (1, -11) of point P into the equation:
-11 = (1/2)(1) + c
-11 = 1/2 + c
To solve for c, subtract 1/2 from both sides:
-11 - 1/2 = c
-22/2 - 1/2 = c
-23/2 = c
Therefore, the equation of the normal to the curve at point P is y = (1/2)x - 23/2.