BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805
BuyFind

Single Variable Calculus: Concepts...

4th Edition
James Stewart
Publisher: Cengage Learning
ISBN: 9781337687805

Solutions

Chapter 3.5, Problem 21E
To determine

To find: The equation of the tangent line to the given curve at the given point.

Expert Solution

Answer to Problem 21E

The equation of the tangent line to the curve ysin2x=xcos2y at (π2,π4) is. y=12x.

Explanation of Solution

Given:

The curve is ysin2x=xcos2y.

The point is (π2,π4).

Derivative rules:

Chain rule

If y=f(u) and u=g(x) are both differentiable function, then dydx=dydududx.

Formula used:

The equation of the tangent line at (x1,y1) is, yy1=m(xx1) (1)

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

Calculation:

Consider ysin2x=xcos2y.

Differentiate the above equation implicitly with respect to x,

yddx(sin2x)+sin2xddx(y)=xddx(cos2y)+cos2yddx(x)yddx(sin2x)+sin2xdydx=xddx(cos2y)+cos2y(1)

Apply the chain rule (1) and simplify the terms,

y[dd(2x)(sin2x)ddx(2x)]+sin2xdydx=x[dd(2y)(cos2y)ddx(2y)]+cos2yy[cos2x(2)]+sin2xdydx=x[2sin2ydydx]+cos2y2ycos2x+sin2xdydx=2xsin2ydydx+cos2y

Combine the terms of dydx,

sin2xdydx+2xsin2ydydx=cos2y2ycos2x(sin2x+2xsin2y)dydx=cos2y2ycos2xdydx=cos2y2ycos2xsin2x+2xsin2y

Therefore, the derivative of y is dydx=cos2y2ycos2xsin2x+2xsin2y.

The slope of the tangent line at (π2,π4) is computed as follows,

m=dydx|(x,y)=(π2,π4)=cos2(π4)2(π4)cos2(π2)sin2(π2)+2(π2)sin2(π4)=cosπ2(π2)cosπsinπ+πsinπ2

Substitute the values cosπ=1 and sinπ2=1,

m=0π2(1)0+π1=π2π=π2×1π=12

Thus, the slope of the tangent line at (π2,π4) is m=12.

Substitute (π2,π4) for (x1,y1) and m=12 in equation (1),

yπ4=12(xπ2)yπ4=12xπ4y=12xπ4+π4y=12x

Therefore, the equation of the tangent line to the curve ysin2x=xcos2y at (π2,π4) is y=12x.

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