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Wyatt Flores

Answered question

2022-04-22

How to determine the equation of the tangent line to

y^{3} + x y = 22

at (7,2)?

dagars5nx

Beginner2022-04-23Added 12 answers

Observe that (7,2) lies on the curve

y^{3} + x y = 22

, as putting these values in the equation

y^{3} + x y = 22

the equality is established as

2^{3} + 7 \times 2 = 8 + 14 = 22

Now, as slope of tangent at any point is given by the value of first derivative

\frac{d y}{d x}

at
that point. Hence, we need to first find

\frac{d y}{d x}

using implicit differentiation.
As

y^{3} + x y = 22

3 y^{2} \frac{d y}{d x} + y + x \frac{d y}{d x} = 0

and.putting x=7 and y=2,

3 \cdot 2^{2} \cdot \frac{d y}{d x} + 2 + 7 \frac{d y}{d x} = 0

\frac{d y}{d x} (12 + 7) = - 2

i.e.

\frac{d y}{d x} = - \frac{2}{19}

As slope of tangent is

- \frac{2}{19}

and it passes through (7,2), its equation using point-slope form is

y - 7 = - \frac{2}{19} (x - 2)

i.e. 2x+19y=137

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