# a) Find the gradient of the function  y=9x-x^3 at the point where x=-3. (b) Use your answer from part (a) to find the equation of:  i.        The tangent to the curve at x= -3...

a) Find the gradient of the function  y=9x-x^3 at the point where x=-3.

(b) Use your answer from part (a) to find the equation of:

i.        The tangent to the curve at x= -3

ii.       The normal to the curve at  x= -3

krishna-agrawala | Student

Gradient of the function y = 9x - x^3 at any point x is given by derivative of this function.

Derivative of 9x - x^3 is:

9 - 3x^2

Gradient of the original function when x = -3 is obtained by substituting the value -3 for x in above derivative. Thus

Gradient at x = -3 is:

= 9 - [3*(-3)^2] = -18

(b) i. Finding equation of tangent to the curve at x = -3

When x = -3, value of y for the curve is obtained substituting this value of x in the equation for the curve. Therefore:

y = 9*(-3) - (-3)^3 = 0

Therefore we have to find the equation of the line that has slope of 36 and passes through the point (-3,0), and has gradient of -18. Therefore equation of the tangent is:

y = -18x + c ... (1)

To find the value of c substituting the coordinates x = 3 and y = 0 in equation (1) we get:

0 = -18*(-3) + c = 54 + c

Therefore c = 18*3 = -54

Substituting this value of c in equation 1, equation of tangent to the curve becomes:

y = -18x - 54

(b) ii. Finding equation of tangent to the curve at x = -3

Gradient of the tangent is -18. Therefore gradient of the normal will be:

= -1/(-18) = 1/18

Therefore we have to find the equation of a line with slope of 1/18 and passing through the point (-3,0).

Thus the equation of the normal is:

y = (1/18)x + c

Or

y = x/18 + c  ...    (2)

To find value of c, substituting the coordinates x = -3 and y = 0 in equation (2) we get:

0 = -3/18 + c = -1/6 + c

Therefore c = 1/6

Substituting this value of c in equation 2, equation of normal to the curve becomes:

y = x/18 + 1/6

neela | Student

The gradient or the slope of the curve, y=f(x)  at x=-3 is the value of the differential coefficient of that function for x=-3.When x= -3 y value of the curve ,9x-x^3 is 9(-3)-(-3)^3 = 0

dy/dx = d/dx of (9x-x^3) =9-3x^2

Therefore,

(d/dx) of (9x-x^3) at x= 3 is  value of (9-3x^2) for x=-3 = 9-3*(-3)^2= -18.

Therefore, -18 is the slope or gradient of the tangent to the curve, y= 9x-x^3 at x=3  and y=0 and  -(-1/18) = (1/18) is the slope of the normal at x = -3 and y= 0

The y value at x=-3 for the given curve,9x-x^3 is 9(-3)-(-3)^3 = 0

Therefore, the equation of tangent at x=-3 and y=0 is

y-0 = (gradient of tangent found above)*[x-(-3)] or

y= -18(x+3) or 18x+y+54 = 0 is the required equation of the tangent.

The equation of the normal is:

y-0 =(gradient of the normal found above)*[x-(-3)] or

y= (1/18)(x+3) or

18y=(x+3) or

x-18y+3 = 0 is the equaion of the normal.