Find the points on the curve y = x3 at which the slope of the tangent is equal to the Y-coordinate of the point.
The equation of the given curve is y = x3
\(\frac{dy}{dx}\)=3x2
The slope of the tangent to a curve at (x, y) is given by,
Therefore, the slope of the tangent at the point where x = 2 is given by,
dy/dx]x,y) =3x2
When the slope of the tangent is equal to the y-coordinate of the point, then y = 3x2 . Also, we have y = x3 .
∴3x2 = x3
∴ x2 (x − 3) = 0
∴ x = 0, x = 3
When x = 0, then y = 0 and when x = 3, then y = 3(3) 2 = 27.
Hence, the required points are (0, 0) and (3, 27).
If \( x = a(0 - \sin \theta) \), \( y = a(1 + \cos \theta) \), find \[ \frac{dy}{dx}. \]
Find the least value of ‘a’ for which the function \( f(x) = x^2 + ax + 1 \) is increasing on the interval \( [1, 2] \).
If f (x) = 3x2+15x+5, then the approximate value of f (3.02) is
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is:
m×n = -1