Question

45-50. tangent lines carry out the following steps.
a. verify that the given point lies on the curve.
b. determine an equation of the line tangent to the curve at the given point.

1. sin y + 5x = y²; (0, 0)
2. x³ + y³ = 2xy; (1, 1)

Explanation:

Step1: Verify point on curve for 45

Substitute \(x = 0\) and \(y = 0\) into \(\sin y+5x=y^{2}\).
\(\sin(0)+5\times0=0^{2}\), since \(\sin(0) = 0\), \(0 + 0=0\), so \((0,0)\) lies on the curve.

Step2: Differentiate implicitly for 45

Differentiate \(\sin y+5x=y^{2}\) with respect to \(x\).
\(\cos y\cdot y'+5 = 2y\cdot y'\).

Step3: Solve for \(y'\) for 45

Rearrange to get \(y'\) terms on one - side: \(\cos y\cdot y'-2y\cdot y'=-5\).
Factor out \(y'\): \(y'(\cos y - 2y)=-5\), so \(y'=\frac{5}{2y - \cos y}\).

Step4: Find slope at the point for 45

Substitute \(x = 0\) and \(y = 0\) into \(y'\): \(y'\big|_{(0,0)}=\frac{5}{2\times0-\cos(0)}=- 5\).

Step5: Find tangent - line equation for 45

Use the point - slope form \(y - y_{1}=m(x - x_{1})\) with \((x_{1},y_{1})=(0,0)\) and \(m=-5\).
\(y-0=-5(x - 0)\), so \(y=-5x\).

Step6: Verify point on curve for 46

Substitute \(x = 1\) and \(y = 1\) into \(x^{3}+y^{3}=2xy\).
\(1^{3}+1^{3}=2\times1\times1\), \(1 + 1=2\), so \((1,1)\) lies on the curve.

Step7: Differentiate implicitly for 46

Differentiate \(x^{3}+y^{3}=2xy\) with respect to \(x\).
\(3x^{2}+3y^{2}y'=2y + 2xy'\).

Step8: Solve for \(y'\) for 46

Rearrange: \(3y^{2}y'-2xy'=2y - 3x^{2}\).
Factor out \(y'\): \(y'(3y^{2}-2x)=2y - 3x^{2}\), so \(y'=\frac{2y - 3x^{2}}{3y^{2}-2x}\).

Step9: Find slope at the point for 46

Substitute \(x = 1\) and \(y = 1\) into \(y'\): \(y'\big|_{(1,1)}=\frac{2\times1-3\times1^{2}}{3\times1^{2}-2\times1}=\frac{2 - 3}{3 - 2}=-1\).

Step10: Find tangent - line equation for 46

Use the point - slope form \(y - y_{1}=m(x - x_{1})\) with \((x_{1},y_{1})=(1,1)\) and \(m=-1\).
\(y - 1=-1(x - 1)\), \(y-1=-x + 1\), so \(y=-x+2\).

Answer:

For \(45\):
a. The point \((0,0)\) lies on the curve \(\sin y + 5x=y^{2}\).
b. The equation of the tangent line is \(y=-5x\).
For \(46\):
a. The point \((1,1)\) lies on the curve \(x^{3}+y^{3}=2xy\).
b. The equation of the tangent line is \(y=-x + 2\).