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

Answer:

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 ,

$y=(x-a) \sqrt{(x-b)}$  --- 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

$0=(b+1-a) \sqrt{(b+1-b)}$

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

According to multiple rule of Differentiation,

$y^{\prime}=u^{\prime} y+y^{\prime} u$

so, we get

${u}^{\prime}=1$

$v^{\prime}=\frac{1}{2} \sqrt{(x-b)}$

$y^{\prime}=1 \times \sqrt{(x-b)}+(x-a) \times \frac{1}{2} \sqrt{(x-b)}$

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

$y^{\prime}=\sqrt{(b+1-b)}+(b+1-a) \times \frac{1}{2} \sqrt{(b+1-b)}$

$y^{\prime}=\frac{d y}{d x}=1$

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