tangent line

Practise tangent line with an original VectorPaths question, worked explanation and related VCE Methods practice.

4 practice marks

Original practice question

The line $y = ax + 2$ is a tangent to the curve $y = x^{\frac{1}{3}} + d$ at the point $(8, c)$ where $a$, $c$ and $d$ are real constants. Find the values of $a$, $c$ and $d$.

Worked explanation

• Find the derivative

  \[\frac{dy}{dx} = \frac{1}{3}x^{-\frac{2}{3}}\]

• Gradient at tangent point

  \[a = \frac{1}{3} \cdot 8^{-\frac{2}{3}} = \frac{1}{3} \cdot \frac{1}{4} = \frac{1}{12}\]

• Find c and d by substitution

  \[c = \frac{1}{12}(8) + 2 = \frac{8}{3}, \quad d = \frac{8}{3} - 8^{\frac{1}{3}} = \frac{8}{3} - 2 = \frac{2}{3}\]

$a = \frac{1}{12}, \; c = 2\frac{2}{3}, \; d = \frac{2}{3}$

Original VectorPaths practice; not an official VCAA question or marking rubric.