MHB Sameer's derivative problem via Facebook

[Prove It]

For the curve with equation $\displaystyle \begin{align*} y = \frac{5}{3}\,x + k\,x^2 - \frac{8}{9}\,x^3 \end{align*}$, calculate the possible values of $\displaystyle \begin{align*} k \end{align*}$ such that the tangents at the points where $\displaystyle \begin{align*} x = 1 \end{align*}$ and $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ are perpendicular

We first need to remember that two lines are perpendicular when their gradients multiply to give -1.

Now, the gradients of the tangents at the points where $\displaystyle \begin{align*} x = 1 \end{align*}$ and $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ can be found by evaluating the derivative at those points.

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{5}{3} + 2\,k \,x - \frac{8}{3}\,x^2 \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} | _{x = 1} &= \frac{5}{3} + 2\,k \cdot 1 - \frac{8}{3} \cdot 1^2 \\ &= 2\,k - 1 \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} |_{x = -\frac{1}{2}} &= \frac{5}{3} + 2\,k \cdot \left( -\frac{1}{2} \right) - \frac{8}{3} \cdot \left( -\frac{1}{2} \right) ^2 \\ &= 1 - k \end{align*}$

Now for the two tangents to be perpendicular, these gradients must multiply to -1...

$\displaystyle \begin{align*} \left( 2\,k - 1 \right) \left( 1 - k \right) &= -1 \\ 2\,k - 2\,k^2 - 1 + k &= -1 \\ 3\,k - 2\,k^2 &= 0 \\ k \left( 3 - 2\,k \right) &= 0 \\ k = 0 \textrm{ or } k &= \frac{3}{2} \end{align*}$

 

[chwala]

We first need to remember that two lines are perpendicular when their gradients multiply to give -1.

Now, the gradients of the tangents at the points where $\displaystyle \begin{align*} x = 1 \end{align*}$ and $\displaystyle \begin{align*} x = -\frac{1}{2} \end{align*}$ can be found by evaluating the derivative at those points.

$\displaystyle \begin{align*} \frac{\mathrm{d}y}{\mathrm{d}x} &= \frac{5}{3} + 2\,k \,x - \frac{8}{3}\,x^2 \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} | _{x = 1} &= \frac{5}{3} + 2\,k \cdot 1 - \frac{8}{3} \cdot 1^2 \\ &= 2\,k - 1 \\ \\ \frac{\mathrm{d}y}{\mathrm{d}x} |_{x = -\frac{1}{2}} &= \frac{5}{3} + 2\,k \cdot \left( -\frac{1}{2} \right) - \frac{8}{3} \cdot \left( -\frac{1}{2} \right) ^2 \\ &= 1 - k \end{align*}$

Now for the two tangents to be perpendicular, these gradients must multiply to -1...

$\displaystyle \begin{align*} \left( 2\,k - 1 \right) \left( 1 - k \right) &= -1 \\ 2\,k - 2\,k^2 - 1 + k &= -1 \\ 3\,k - 2\,k^2 &= 0 \\ k \left( 3 - 2\,k \right) &= 0 \\ k = 0 \textrm{ or } k &= \frac{3}{2} \end{align*}$

This is correct! One can also verify by finding the equations of the tangent. For instance, when $k=0$ then our tangent equations would be, $y=x-0.222$ and $y=-x+1.778$ that coincide with each other at $90^0$. Similarly this can be done for $k=\dfrac{3}{2}$.