Question

If the normal to the curve $y\left(x\right)={\int }_0^x\left(2t^2-15t+10\right)dt$ at a point $\left(a,b\right)$ is parallel to the line $x+3y=-5$ and it is also given that $a>1$, then the value of $\left|a+6b\right|$ is equal to _______.

Answer 1

Step 1: Calculate the slope of the curve

The given equation of the curve $y\left(x\right)={\int }_0^x\left(2t^2-15t+10\right)dt$.

Differentiate both sides of the equation with respect to $x$.

$$\begin{gathered} \frac{d}{dx}\left[y\left(x\right)\right]=\frac{d}{dx}\left[{\int }_0^x\left(2t^2-15t+10\right)dt\right] \\ \Rightarrow y\text{'}\left(x\right)=\left(2x^2-15x+10\right)\frac{d\left(x\right)}{dx}-\left(2{\left(0\right)}^2-15\left(0\right)+10\right)\frac{d\left(0\right)}{dx} \\ \Rightarrow y\text{'}\left(x\right)=\left(2x^2-15x+10\right)-0 \\ \Rightarrow y\text{'}\left(x\right)=2x^2-15x+10 \end{gathered}$$

Thus, the slope of the normal at a point $\left(a,b\right)$ can be given by $m_1=-\frac{1}{y\text{'}\left(a\right)}$.

$\Rightarrow m_1=-\frac{1}{2a^2-15a+10}$.

Step 2: Calculate the value of $a$

The normal is parallel to the line $x+3y=-5$.

$\Rightarrow y=-\frac{1}{3}x-\frac{5}{3}$

Compare the given equation of the straight line with the slope-intercept form of a straight line, $y=mx+c$, where $m$ is the slope.

Thus, the slope of the line is $m_2=-\frac{1}{3}$.

As the lines are parallel, thus $m_1=m_2$.

$$\begin{gathered} \Rightarrow -\frac{1}{2a^2-15a+10}=-\frac{1}{3} \\ \Rightarrow 2a^2-15a+10=3 \\ \Rightarrow 2a^2-15a+7=0 \\ \Rightarrow 2a^2-14a-a+7=0 \\ \Rightarrow 2a\left(a-7\right)-1\left(a-7\right)=0 \\ \Rightarrow \left(2a-1\right)\left(a-7\right)=0 \\ \Rightarrow a=\frac{1}{2},7 \end{gathered}$$

It is given that, $a>1$.

So, $a=7$.

Step 3: Calculate the value of $b$

The given point $\left(a,b\right)$ also lies on the curve $y\left(x\right)={\int }_0^x\left(2t^2-15t+10\right)dt$

Thus, $y\left(a\right)={\int }_0^a\left(2t^2-15t+10\right)dt$

$$\begin{gathered} \Rightarrow b={\int }_0^7\left(2t^2-15t+10\right)dt\left[\because a=7\right] \\ \Rightarrow b={\left[\frac{2t^3}{3}-\frac{15t^2}{2}+\frac{10t}{1}\right]}_0^7 \\ \Rightarrow b=\left[\frac{2{\left(7\right)}^3}{3}-\frac{15{\left(7\right)}^2}{2}+\frac{10\left(7\right)}{1}-\frac{2{\left(0\right)}^3}{3}+\frac{15{\left(0\right)}^2}{2}-\frac{10\left(0\right)}{1}\right] \\ \Rightarrow b=\frac{686}{3}-\frac{735}{2}+70-0+0-0 \\ \Rightarrow b=\frac{\left(686\times 2\right)-\left(735\times 3\right)+\left(70\times 6\right)}{6} \\ \Rightarrow 6b=-413 \end{gathered}$$

Therefore, $\left|a+6b\right|=\left|7-413\right|$.

$\Rightarrow \left|a+6b\right|=406$.

Hence, the value of $\left|a+6b\right|$ is $406$.