Question

Find the slope of the tangent of the curve y² - 3x³ + 2 = 0 at (1, -1)
1. -1.5
2. 2.5
3. 3.5
4. -4.5

Answer 1

Correct Answer - Option 4 : -4.5

Concept:

The slope of the tangent to a curve y = f(x) is m = \(\rm dy\over dx\)

The slope of the normal = \(\rm -{1\over m}\) = \(\rm -{1\over {dy\over dx}}\)

Calculation:

Given curve y2 - 3x3 + 2 = 0

Differentiating the equation wrt x we get

⇒ 2y\(\rm dy\over dx\) - 9x2 + 0 = 0

⇒ 2y \(\rm dy\over dx\)  = 9x2

Slope at (1, -1)

⇒ 2(-1) \(\rm dy\over dx\) = 9 (1)

⇒ -2\(\rm dy\over dx\) = 9

⇒ \(\rm dy\over dx\) = -4.5

The slope of the tangent (m) = \(\rm {dy\over dx}\)

∴ m = -4.5