#### To determine

**To find:** The equation of the tangent line to the curve which is parallel to another line.

#### Answer

The equation of tangent line to the curve that is parallel to the line 32x−y=15 is y=32x−47_.

#### Explanation

**Given:**

The curve is y=x4+1 and the line is 32x−y=15.

**Derivative rules:**

(1) Derivative of constant function: ddx(c)=0

(2) Power Rule: ddx(xn)=nxn−1

(3) Sum rule: ddx(f+g)=ddx(f)+ddx(g)

**Formula Used:**

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

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

**Calculation:**

**Step 1:**

Obtain the slope of the tangent line to the curve.

The derivative of curve *y* is as follows,

dydx=ddx(x4+1)

Apply the derivative rule (1) and the sum rule (3).

dydx=ddx(x4)+ddx(1)=ddx(x4)+0

Apply the power rule (2) and simplify the expression.

dydx=4x4−1=4x3

Thus, the derivative of curve *y* is 4x3.

Since the derivative of the curve is the slope of the tangent line to the curve.

Therefore, the slope of the tangent line to the curve y=x4+1 is 4x3.

**Step 2:**

Obtain the tangent point.

Express the given line as,y=32x−15 and its slope is 32.

Since the required equation of the tangent line is parallel to the given line, the slope of the tangent line is equal to the slope of the given line.

That is, 4x3=32.

Solve the equation 4x3=32 and obtain the value of *x* as 2.

Substitute 2 for *x* in y=x4+1 and obtain the value of *y*.

y=x4+1=24+1=16+1=17

Thus, the value of y=17.

Therefore, the tangent point is (2,17).

**Step 3:**

Obtain the equation of the tangent line to the curve.

Substitute (2,17) for (x1,y1) and 32 for *m* in equation (1),

y−17=32(x−2)y−17=32x−64

Add 17 on both sides,

y=32x−64+17y=32x−47

Therefore, the equation of the tangent line to the curve that is parallel to the line 32x−y=15 is y=32x−47.