#### To determine

**To find:** The equation of the tangent line and the normal line to the curve at the point.

#### Answer

The equation of the tangent line to the curve y2=x3 at (1,1) is y=32x−12.

The equation of the normal line to the curve y2=x3 at (1,1) is y=−23x+53_.

#### Explanation

**Given:**

The curve is y2=x3.

The curve passing through the point is (1,1).

**Formula used:**

Power Rule: ddx(xn)=nxn−1

**Formula used:**

The equation of the tangent line at (x1,y1) is, y−y1=m(x−x1) (1)

The equation of normal line at (x1,y1) is, y−y1=−1m(x−x1) (2)

Here, *m* is the slope of the tangent line at (x1,y1) and m=dydx|x=x1.

**Calculation:**

The given equation y2=x3 can be written as follows,

y2=x3y=(x3)12y=x32

The derivative of y is dydx, which is obtained as follows.

dydx=ddx(y) =ddx(x32)

Apply power rule and simplify the expressions.

dydx=32x32−1=32x3−22=32x12

Therefore, the derivative of the function y2=x3 is 32x12_.

The slope of the tangent line at (1,1) is,

m=dydx|x=1=32⋅(1)12=32⋅1=32

Thus, the slope of the tangent line at (1,1) is m=32.

Substitute (1,1) for (x1,y1) and 32 for *m* in equation (1),

y−1=32(x−1)y−1=32x−32

Add 1 on both sides and simplify the terms.

y=32x−32+1y=32x−32+22y=32x−12

Therefore, the equation of the tangent line to the curve y2=x3 at (1,1) is y=32x−12.

Substitute (1,1) for (x1,y1) and 32 for *m* in equation (1),

y−1=−23(x−1)y−1=−2x3+23

Add 1 on both sides and simplify the terms.

y=−2x3+23+1y=−2x3+23+33y=−2x3+53

Therefore, the equation of the normal line to the curve y2=x3 at (1,1) is y=−23x+53_.