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find the equation of the tangent to the curve at x=3 for parametric equations

x = t + 1/t
y = t^2 + 1/(t^2) when t is greater than 0

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MathPhys

Answer:

y = 6x − 11

Step-by-step explanation:

x = t + (1/t), y = t² + (1/t²)

If we square x:

x² = t² + 2 + (1/t²)

x² = y + 2

When x = 3, y = 7.

Taking derivative with respect to time:

2x = dy/dx

dy/dx = 6

So the equation of the tangent line is:

y − 7 = 6 (x − 3)

y − 7 = 6x − 18

y = 6x − 11

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