# Tutor profile: Harish J.

## Questions

### Subject: Pre-Calculus

What is the equation of the line tangent to the curve $$ y = 3e^{2x} $$ and perpendicular to the line $$ y = -2x + 10 $$

The slope of line $$ y = -2x + 10 $$ is $$ m_1 = -2 $$. It is given that the tangent line is perpendicular to this line. Hence, if $$ m_2 $$ is the slope of the tangent line, then we have: $$ m_1 m_2 = -1 $$ or $$ -2 m_2 = -1 $$ or $$ m_2 = \frac{1}{2} $$ Hence, the tangent line belongs to the family of lines: $$ y = \frac{1}{2}x + b $$ Now, a tangent to the curve $$ y = 3e^{2x} $$ at a general point $$ (x,y) $$ would have slope = $$ \frac{dy}{dx} = 6e^{2x} $$. Hence, $$ 6e^{2x} = \frac{1}{2} $$ or $$ e^{2x} = \frac{1}{12} $$ or $$ x = -1.24245 $$ Hence, the tangent line intersects the curve at $$ x = -1.24245, y = 3e^{2x} = 0.25 $$. Its slope is $$ m_2 = \frac{1}{2} $$. Hence, the required equation of the tangent line is: $$ y - 0.25 = \frac{1}{2} (x - (-1.24245)) $$ or $$ y = \frac{1}{2}x + 0.871 $$

### Subject: Calculus

Find the derivative $$ \frac{\mathrm{d}y}{\mathrm{d}x}$$: $$ y^3 + x^3 = 3(y-x) $$

Using implicit differentiation, we have: $$ 3y^2\frac{\mathrm{d}y}{\mathrm{d}x} + 3x^2 = 3( \frac{\mathrm{d}y}{\mathrm{d}x} - 1)$$ or $$ (3y^2-3) \frac{\mathrm{d}y}{\mathrm{d}x} = -x^3 -x$$ or $$ \frac{\mathrm{d}y}{\mathrm{d}x} = \frac{x^3 + x}{3-3y^2} $$

### Subject: Algebra

Find the equation of a polynomial with roots 2 and -5i, real coefficients, and passing through the point (1, 52).

The polynomial has a complex root x = -5i. Hence, it also has the conjugate root x = +5i, so that it would have real coefficients. The root x = 2 is also specified making a total of minimum 3 roots. So, we can choose a cubic polynomial from the family: $$ f(x) = k(x - 2)(x - (-5i))(x - 5i) $$ or $$ f(x) = k(x - 2)(x^2+25) $$ It is given that it passes through (1, 52). Hence, we have: $$52 = k(1-2)(1^2+25)$$ or $$ k = -2$$ Hence, the required polynomial is: $$ f(x) = -2(x-2)(x^2+25)$$ or $$ f(x) = -2x^3+4x^2-50x+100 $$

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