Calculus Question

kalyan601

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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!
 
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.

I don't know what you mean by "both y and x= 8". You were told that x= 4, not 8.
To prove that (4, 8) is on C, show that 4(4)+ 3(43/2)- 2(42)= 16+ 3(8)- 2(16)= 16+ 24- 32= 40- 32= 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.
Yes, that is correct.

Normal Gradient is negative reciprocal of tangent gradient.
y= x/3 + c
sub (4,8), 3y = x+ 20
Setting x= 4, y= 8, 8= 4/3+ c so c= 8- 4/3= (24- 4)/3= 20/3 so y= x/3+ 20/3. Multiplying on both sides by 3: Yes, 3y= x+ 20.

d) By Pythagoras, a2= 16+64 = 80. Distance PQ = square root of 80.
This is the distance from P to (0, 0). How did you determine that Q is (0, 0)?
The line 3y= x+ 20 does NOT cut the x-axis at (0, 0).
 
Last edited:
"The line 3y= x+ 20 does NOT cut the x-axis at (0, 0)."

Hint for the OP: It cuts the x-axis when y = 0.
 
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