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Tangents & Normals (1 Viewer)

mtsmahia

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Hi guys, I had 2 Q in differentiation.

For the curve y=ax^2 + bx + c, where a, b, c are constants, it is given that at the points (2,12) and (-1, 0), the slope of the tangent is 7 & 1 respectively. Find a,b & c.

Im stuck at the simulataneous eq part...

The other Q is

Find the eqns of the normal at the points where the curve y=x^2(x^2-1) cuts the x-axis.


thanksfor the help!!!
 
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lyounamu

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Hi guys, I had 2 Q in differentiation.

For the curve y=ax^2 + bx + c, where a, b, c are constants, it is given that at the points (2,12) and (-1, 0), the slope of the tangent is 7 & 1 respectively. Find a,b & c.

Im stuck at the simulataneous eq part...

The other Q is

Find the eqns of the normal at the points where the curve y=x^2(x^2-1) cuts the x-axis.


thanksfor the help!!!

1st Q:
dy/dx = 2ax + b
when x = 2, dy/dx = 7
when x = -1, dy/dx = 1

i.e. 4a + b = 7 (1)
and -2a + b = 1 (2)

(1) - (2)

6a = 6
a = 1
so b = 3

so y = x^2 + 3x + c
if it crosses (2,12)

12 = 4 + 6 + c
c = 2
2nd Q:

Find the eqns of the normal at the points where the curve y=x^2(x^2-1) cuts the x-axis

it crosses x-axis when y = 0 i.e. when x = 0, 1, -1

y = x^4 - x^2
dy/dx = 4x^3 - 2x

when x = 0, dy/dx = 0
when x = 1, dy/dx = 2
when x= -1, dy/dx = -2

i.e. the gradients of normal are -1/2 and 1/2
now use the y-y1 = m(x-x1) to find each equation
 
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mtsmahia

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Just another Q

Find the gradient of the tangent on the curve y=x/square root of (2x) at the point (4, sqaure root of 2) , expressing the anwer rationally.
 

lyounamu

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Just another Q

Find the gradient of the tangent on the curve y=x/square root of (2x) at the point (4, sqaure root of 2) , expressing the anwer rationally.
y=x/square root of (2x)
= sqrt(x)/sqrt(2)

dy/dx = 1/(2sqrt(2x))

when x=4, dy/dx = 1/2sqrt(8) = sqrt(8)/16 = sqrt(2)/8
 

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