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This is one of our many quizzes on Derivative applications and graphing. Click on this to access more such competitive quizzes on Differential Calculus.

Derivative applications and graphing Quiz 2

mathematics

 What point on the function $y={{x}^{2}}-6x+10$ is closest to the origin? 

 Take the point as $\left( x,y \right)$ and find the distance of the point from the origin , substitute $y={{x}^{2}}-6x+10$  and then differentiate and equate to 0 to find the value of $x$and  $y$.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/cf7d944c-e280-4faa-86e4-cd87aeb072193421895515411604987.png"/> $\text{Let the closest point on curve be}\left( x,y \right) \\ \text{Distance from origin , }d=\sqrt{{{x}^{2}}+{{y}^{2}}} \\ \Rightarrow d=\sqrt{{{x}^{2}}+{{\left( {{x}^{2}}-6x+10 \right)}^{2}}} \\ \text{The values of }x\text{ which minimize }d\text{ also minimizes }{{d}^{2}} \\ \Rightarrow {{d}^{2}}={{x}^{2}}+{{\left( {{x}^{2}}-6x+10 \right)}^{2}} \\ \text{To minimize , make the differentiation equals 0} \\ \Rightarrow {{\left( {{d}^{2}} \right)}^{\prime }}=2x+2\left( {{x}^{2}}-6x+10 \right)\left( 2x-6 \right)=0 \\ \Rightarrow 4{{x}^{3}}-36{{x}^{2}}+114x-120=0 \\ \Rightarrow x=2.26 \\ \text{For minima , }{{\left( {{d}^{2}} \right)}^{\prime }}^{\prime }>0 \\ (d^2)’’ =12x^2-72x+114 \\ \therefore x=2.26 \\ \Rightarrow y={{\left( 2.26 \right)}^{2}}-6\left( 2.26 \right)+10 \\ \therefore y=1.55 $

 Find the maximum value of $f\left( x \right)=4{{x}^{3}}-60{{x}^{2}}+216x$

First find ${f}'\left( x \right)$ and the critical points of the function by equating ${f}'\left( x \right)$ to 0 and then findMaxima by using the condition ${{f}'}'\left( x \right)<0$

&nbsp;<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/e68df80e-1628-4f45-a92b-ced65a704331667502174909384630.png"/> $f\left( x \right)=4{{x}^{3}}-60{{x}^{2}}+216x \\ \text{For critical points make the differentiation equal to 0} \\ {f}'\left( x \right)=216-120x+12{{x}^{2}}=0 \\ \Rightarrow {{x}^{2}}-10x+18=0 \\ \Rightarrow x=2.35,7.65 \\ \text{For maxima, the condition is }{{f}'}'\left( x \right)<0\Rightarrow {{f}'}'\left( x \right)=24x-120<0 \\ \Rightarrow {{f}'}'\left( 2.35 \right)=24\left( 2.35 \right)-120<0 \\ \therefore \text{At }x=2.35\text{ we will have maximum value} \\ f\left( 2.35 \right)=4{{\left( 2.35 \right)}^{3}}-60{{\left( 2.35 \right)}^{2}}+216\left( 2.35 \right) \\ \therefore f\left( 2.35 \right)=228.1615 $

 A particle moves along the curve $y={{x}^{2}}+2$. The x-coordinate increases at a rate of $4\,cm/s$. How far is the distance between the point and the origin changing when $x=1$?  

 By using the value of $\dfrac{dx}{dt}$ find the value of $\dfrac{dy}{dt}$, then use formula to find the value of the distance between the point and the origin , then differentiate it to obtain the answer.

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/a33b722b-1487-4b24-853e-6039e7f841d56437235639481984795.png"/> $ \text{Given }x=1\Rightarrow y={{\left( 1 \right)}^{2}}+2=3\,\text{and}\,\dfrac{dx}{dt}=4 \\ y={{x}^{2}}+2 \\ \Rightarrow \dfrac{dy}{dt}=2x\,\dfrac{dx}{dt}=2\left( 1 \right)\left( 4 \right) \\ \Rightarrow \dfrac{dy}{dt}=8 \\ \text{From the graph, we can say that} \\ {{z}^{2}}={{x}^{2}}+{{y}^{2}} \\ \Rightarrow {{z}^{2}}={{\left( 1 \right)}^{2}}+{{\left( 3 \right)}^{2}}=10 \\ \therefore z=\sqrt{10} \\ \text{Now by differentiating }{{z}^{2}}={{x}^{2}}+{{y}^{2}}\text{,} \\ 2z\dfrac{dz}{dt}=2x\dfrac{dx}{dt}+2y\dfrac{dy}{dt} \\ \Rightarrow \sqrt{10}\dfrac{dz}{dt}=4+3\left( 8 \right) \\ \Rightarrow \dfrac{dz}{dt}=\dfrac{28}{\sqrt{10}} \\ \therefore \dfrac{dz}{dt}=\dfrac{14\sqrt{10}}{5}\,cm/s$

 If the minimum value of $g\left( x \right)={{x}^{3}}-9{{x}^{2}}+24x+3$ is $\alpha $. Then find the value of $\sqrt{\alpha -3}$

 First find ${f}'\left( x \right)$ and the critical points of the function by equating ${f}'\left( x \right)$ to 0 and then find Minima by using the condition ${{f}'}'\left( x \right)>0$

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/4be33451-4604-4a07-9fef-db91b85caaf07906842445818905346.png"/> $g\left( x \right)={{x}^{3}}-9{{x}^{2}}+24x+3 \\ \text{For critical points make the differentiation equal to 0} \\ \text{{g}'}\left( x \right)=3{{x}^{2}}-18x+24=0 \\ \Rightarrow {{x}^{2}}-6x+8=0 \\ \Rightarrow x=2,4 \\ \text{For minima, the condition is }{{g}'}'>0\,\Rightarrow \,{{g}'}'\left( x \right)=6x-18>0 \\ \Rightarrow {{g}'}'\left( 4 \right)=6\left( 4 \right)-18>0 \\ \therefore \text{At }x=4\text{ we will have the minimum value} \\ \text{g}\left( 4 \right)={{4}^{3}}-9{{\left( 4 \right)}^{2}}+24\left( 4 \right)+3 \\ \therefore g\left( 4 \right)=19=\alpha \\ \therefore \sqrt{\alpha -3}=\sqrt{19-3}=4 $

 Find the absolute maximum and absolute minimum of the function $f\left( x \right)=2{{x}^{3}}-5{{x}^{2}}+4x-1$ on $\left[ -1,2 \right]$

  First find ${f}'\left( x \right)$ and the critical points on the function by equating ${f}'\left( x \right)$ to 0 and then find absolute minimum and absolute maximum from critical points and the given interval $\left[ -1,2 \right]$

<img src="https://cmds-vedantu.s3.ap-southeast-1.amazonaws.com/prod/question-sets/06177e5a-e4c7-4da0-9ae6-259ee81430698320785803052950073.png"/> $ f\left( x \right)=2{{x}^{3}}-5{{x}^{2}}+4x-1 \\ {f}'\left( x \right)=6{{x}^{2}}-10x+4=2\left( 3x-2 \right)\left( x-1 \right) \\ \text{The critical numbers are the solutions of }{f}'\left( x \right)=0 \\ \Rightarrow x=\dfrac{2}{3},1 \\ \text{Both these critical numbers }\dfrac{2}{3}\text{ and 1 are in }\left( -1,2 \right). \\ f\left( \dfrac{2}{3} \right)=2{{\left( \dfrac{2}{3} \right)}^{3}}-5{{\left( \dfrac{2}{3} \right)}^{2}}+4\left( \dfrac{2}{3} \right)-1=\dfrac{1}{27} \\ f\left( 1 \right)=2{{\left( 1 \right)}^{3}}-5{{\left( 1 \right)}^{2}}+4\left( 1 \right)-1=0 \\ f\left( -1 \right)=2{{\left( -1 \right)}^{3}}-5{{\left( -1 \right)}^{2}}+4\left( -1 \right)-1=-12 \\ f\left( 2 \right)=2{{\left( 2 \right)}^{3}}-5{{\left( 2 \right)}^{2}}+4\left( 2 \right)-1=3 \\ \text{Hence, }3\text{ is the absolute maximum, achieved at }x=2 \\ -12\text{ is the absolute minimum , achieved at }x=-1$