Hi, I've got this question that I have done but am not sure on whether I used the correct methods. Can someone tell me if I have and if not, how I should improve/do it properly.
The curve C has equation y=4x + 3x(3/2) - 2x2 , x>0
a) Find an expression for dy/dx
b) Show that point P(4,8) lies on C
c) Show that an equation of the normal to C at the point P is 3y = x+20
d) The normal to C at P cuts the x-axis at point Q. Find length PQ
a) dy/dx = 4 + (9/2)x(1/2)-4x
b) Sub x=4 and y=8 into C equation. 8=4(4) + 3(4)(3/2) - 2(4)2. Both y and x = 8. Point P lies on C
c) Gradient of C at point P is x=4 substituted into dy/dx.
gradient = 4+ 9-16 = -3.
Normal Gradient is negative reciprocal of tangent gradient.
y= x/3 + c
sub (4,8), 3y = x+ 20
d) By Pythagoras, a2= 16+64 = 80. Distance PQ = square root of 80.
Thanks!
The curve C has equation y=4x + 3x(3/2) - 2x2 , x>0
a) Find an expression for dy/dx
b) Show that point P(4,8) lies on C
c) Show that an equation of the normal to C at the point P is 3y = x+20
d) The normal to C at P cuts the x-axis at point Q. Find length PQ
a) dy/dx = 4 + (9/2)x(1/2)-4x
b) Sub x=4 and y=8 into C equation. 8=4(4) + 3(4)(3/2) - 2(4)2. Both y and x = 8. Point P lies on C
c) Gradient of C at point P is x=4 substituted into dy/dx.
gradient = 4+ 9-16 = -3.
Normal Gradient is negative reciprocal of tangent gradient.
y= x/3 + c
sub (4,8), 3y = x+ 20
d) By Pythagoras, a2= 16+64 = 80. Distance PQ = square root of 80.
Thanks!