area between curve and tangent line

**Tweety** · April 27th 2009, 11:18 AM

Figure 2 shows the curve

C with equation $y = 3x-4\sqrt{x} + 2$ and the tangent to C at the point A.

Given that A has x-coordinate 4,

(a) show that the tangent to C at A has the equation y = 2x -2.

The shaded region is bounded by

C, the tangent to C at A and the positive

coordinate axes.

(b) Find the area of the shaded region.

**Jameson** · April 27th 2009, 12:51 PM

You need to explain in your posts what you know how to do and what you don't. This will help us help you better.

So for (a) you need to find the derivative of your equation and construct a tangent line formula. This is fairly standard, so I'll assume you know about how to do this unless you tell me otherwise.

Once you find this line, you need to use integration to find the area of the shaded region. First, what is the region of integration? Put another way, what are the bounds of integration? When you have an area between two curves, call it h(x), you write it in terms of the two equations. So is the region made by C-C' or C'-C?

**Tweety** · April 28th 2009, 02:06 AM

Originally Posted by Jameson

You need to explain in your posts what you know how to do and what you don't. This will help us help you better.

So for (a) you need to find the derivative of your equation and construct a tangent line formula. This is fairly standard, so I'll assume you know about how to do this unless you tell me otherwise.

Once you find this line, you need to use integration to find the area of the shaded region. First, what is the region of integration? Put another way, what are the bounds of integration? When you have an area between two curves, call it h(x), you write it in terms of the two equations. So is the region made by C-C' or C'-C?

$\frac{dy}{dx} = 3-2x^{-\frac{1}{2}}$

$x = 4 , 3-1=2$

$y -6 = 2(x-4)$
$y = 2x-2$

I still not sure how to work out the area of the shaded region, Should it be curve minus tangent line ?

**mr fantastic** · April 28th 2009, 05:22 AM

Originally Posted by Tweety

$\frac{dy}{dx} = 3-2x^{-\frac{1}{2}}$

$x = 4 , 3-1=2$

$y -6 = 2(x-4)$
$y = 2x-2$

I still not sure how to work out the area of the shaded region, Should it be curve minus tangent line ?

The area of the shaded region is given by $\int_0^1 3x - 4 \sqrt{x} + 2 \, dx + \int_1^4 (3x - 4 \sqrt{x} + 2) - (2x - 2) \, dx$ .

Do you understand why? (Note that the x-intercept of the tangent line is (1, 0))

**Tweety** · April 28th 2009, 10:57 AM

Originally Posted by mr fantastic

The area of the shaded region is given by $\int_0^1 3x - 4 \sqrt{x} + 2 \, dx + \int_1^4 (3x - 4 \sqrt{x} + 2) - (2x - 2) \, dx$ .

Do you understand why? (Note that the x-intercept of the tangent line is (1, 0))

Yes I get now, thanks.

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