Question

1. Find the point on the curve y = x2 that is closest to (0, 5)....

1. Find the point on the curve y = x2 that is closest to (0, 5).

2. Find the function f(x),iff′′(x)=sinx+x and f(0)=f(π)=0.

3. Find derivatives of the following functions. a) arcsin( square root 3x)

1.

y = x2

Let the point on the curve be (a,b)

As this point lies on the curve, it will satisfy the equation of the curve. Therefore,

b = a^2

Therefore, the point is (a, a2)

The slope of the line joining the point on the curve and the point (0,5) is

Slope = (y2-y1)/(x2-x1) = (a^2 - 5)/(a - 0) = (a^2-5)/a

Slope at any point on the curve = dy/dx = 2x

At the point (a,a^2), Slope = 2a

We know that the line connecting the point on the curve and the point (0,5) will be perpendicular to the tangent at the point on the curve.

Therefore, the product of the two slopes will be -1.

Therefore,

((a^2-5)/a)*2a = -1

(a^2-5)*2 = -1

(a^2-5) = -1/2

(a^2) = (-1/2)+5 = 9/2

a = sqrt(9/2)

a = +3/sqrt(2) and a = -3/sqrt(2)

Therefore, there are 2 points which are closest to the point.

(3/sqrt(2), 9/2) and (3/sqrt(2),9/2)

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