Question

A curve is given by y=(x-a)√(x-b) for x≥b, where a and b are constants, cuts the x axis at A where x=b+1. Show that the gradient of the curve at A is 1.

Answer 1

A curve is given by y=(x-a)√(x-b) for x≥b. The gradient of the curve at A is 1.

Solution:

We need to show that the gradient of the curve at A is 1

Here given that ,

`y=(x-a) √((x-b))` --- equation 1

Also, according to question at point A (b+1,0)

So curve at point A will, put the value of x and y

`0=(b+1-a) √((b+1-b))`

0=b+1-c --- equation 2

According to multiple rule of Differentiation,

`y^(\prime)=u^(\prime) y+y^(\prime) u`

so, we get

`{u}^(\prime)=1`

`v^(\prime)=(1)/(2) √((x-b))`

`y^(\prime)=1 * √((x-b))+(x-a) * (1)/(2) √((x-b))`

By putting value of point A and putting value of eq 2 we get

`y^(\prime)=√((b+1-b))+(b+1-a) * (1)/(2) √((b+1-b))`

`y^(\prime)=(d y)/(d x)=1`

Hence proved that the gradient of the curve at A is 1.