Posted by Amponsah on Monday, April 9, 2012 at 4:35am.
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 xaxis. Find
i) the values of a and b
ii) the equation of the curve

calculus  Reiny, Monday, April 9, 2012 at 9:50am
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
Answer This Question
Related Questions
 Calculus  Maths  I got a few questions. Hope ya'll can help out. 1) for F(X...
 Maths  a curve ahs parametric equations x=t^2 and y= 11/2t for t>0. i)find ...
 calculus  Suppose y is defined implicitly as a function of x by x^2+Axy^2+By^3=...
 calculus  Consider line segments which are tangent to a point on the right half...
 Calculus  Consider line segments which are tangent to a point on the right half...
 math  (a) Integrate wrt x:(〖2x〗^4 x^2+2)/x^2 (b) The gradient of...
 Calculus  Damon  Find the line which passes through the point (0, 1/4) and is ...
 Maths  I need some help with my maths questions please! What angle does the ...
 math  With regards to question J: The variables x and y are connected by the ...
 CalculasDifferentiation  The gradient of the tangent to a curve y=ax^2+bx at ...
More Related Questions