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Related rates Quiz 2

mathematics

 Find $f^\prime (x)$ if, $f(x)={{r}^{2}}+r\sin r-\dfrac{2\sqrt{r}}{5}$.

 To find $f^\prime (x)$ we differentiate the given equation with respect to $x$ and applying the product rule for differentiation we find the value of $f^\prime (x)$.

  $ f(x)={{r}^{2}}+r\sin r-\dfrac{2\sqrt{r}}{5} \\ \text{Differentiating both sides w.r.t }x, \\ r = \dfrac{dr}{dx} \\ f^\prime (x)=2rr^\prime +r^\prime \sin r+rr^\prime \cos r-\dfrac{r^\prime }{5\sqrt{r}} $ 

 For a  function  $2{{y}^{3}}+5{{x}^{2}}=7y+{{e}^{{{y}^{2}}-4y}}$,  given that $x=2$,$y=4$ and $y^\prime =2$. Find  $x^\prime $.

 First differentiate the given equation partially wrt to t where we take $y = \dfrac{dy}{dt} , x^\prime =\dfrac{dx}{dt}$ and then put the values given in the question to find the value of $x^\prime $.

  $2{{y}^{3}}+5{{x}^{2}}=7y+{{e}^{{{y}^{2}}-4y}} \\ \text{On differentiating partially wrt }t\text{ where }y = \dfrac{dy}{dt} \\ x^\prime =\dfrac{dx}{dt} \\ 6{{y}^{2}}y^\prime +10xx^\prime =7y^\prime +{{e}^{{{y}^{2}}-4y}}(2y-4)y^\prime \\ \text{Put the given values from the question} \\ 6{{(4)}^{2}}(2)+10(2)x^\prime =7(2)+{{e}^{8-8}}(2(4)-4)2 \\ 192+20x^\prime   = 14+(16-8) \\20x^\prime  =-170\\ x^\prime  =-8.5 $ 

 If $f(x)={{r}^{2}}+2rh+1$  and  $f(y)=2{{r}^{2}}+4rh+2{{h}^{2}}$. Then find  $\dfrac{f^\prime (x)}{f^\prime (y)}$ when  $r^\prime =2$ and $h^\prime =1$.

 Differentiate the first function w.r.t $x$ and the second function w.r.t $y$ and then pout the value of $r^\prime $ and $h$; and then find the ratio of the functions

  $ f(x)={{r}^{2}}+2rh+1 \\ f^\prime (x)=rr^\prime +2(r^\prime h+rh^\prime ) \\ f(y)=2{{r}^{2}}+4rh+2{{h}^{2}} \\ f^\prime (y)=4rr^\prime +4(r^\prime h+rh^\prime )+4hh^\prime  \\ \text{If }r = 2, h^\prime  =1 \\ f^\prime (x) = 4r+2(2h+r) \\ f^\prime (y)=8r+4h+4(2h+r) \\ \dfrac{f^\prime (x)}{f^\prime (y)}=\dfrac{2(2r+2h+r)}{4(3r+5h)} \\ \Rightarrow \dfrac{3r+2h}{2(3r+5h)} $ 

 Solve for $x$, given   ${{x}^{2}}+xy-{{y}^{2}}=\log x$ , for  $x^\prime =2$,$y=2$ ,$y^\prime =4$.

 Differentiate the above equation w.r.t to $y$ and then put the values of $x^\prime , y, y^\prime $ and then find the value of $x$ which can be obtained by solving the quadratic equation.

  $ {{x}^{2}}+xy-{{y}^{2}}=\log x \\ \text{Differentiating partially wrt }y\text{ where }x = \dfrac{dx}{dt},y^\prime =\dfrac{dy}{dt} \\ \text{2x}\dfrac{dx}{dy}+x+y\dfrac{dx}{dy}-2y=\dfrac{1}{x}\dfrac{dx}{dy} \\ 2{{x}^{2}}\dfrac{x^\prime }{y^\prime }+{{x}^{2}}+xy\dfrac{x^\prime }{y^\prime }-2xy=\dfrac{x^\prime }{y^\prime } \\ \text{Put the values given in question} \\ 2{{x}^{2}}\times \dfrac{2}{4}+{{x}^{2}}+2x\times \dfrac{2}{4}-2x\times 2=\dfrac{2}{4} \\ {{x}^{2}}-x=\dfrac{1}{4} \\ x=1\pm \dfrac{1}{\sqrt{2}} $ 

 Find the product of the rate of change of $r$ with respect to time $t$ in the equation  ${{r}^{3}}+2rh+{{h}^{2}}$ and the reciprocal of  ${{r}^{3}}+{{\left( r-h \right)}^{2}}$ for  $r=3$, $h=4$ and  $h^\prime =2$.

 Consider both the equations and differentiate them w.r.t to $t$ and then find the product of $\dfrac{dr}{dt}$ in both the equations by taking reciprocal of it in the second equation.

  $ {{r}^{3}}+2rh+{{h}^{2}} \\ \text{Differentiating  the above equation partially wrt }t\text{ we get:} \\ \text{3}{{\text{r}}^{2}}\dfrac{dr}{dt}+2\dfrac{dr}{dt}h+2r\dfrac{dh}{dt}+2h\dfrac{dh}{dt}=0 \\ \dfrac{dr}{dt}=\dfrac{-2(rh^\prime +hr^\prime )}{3{{r}^{2}}+2h} \\ \text{Now put the value of }r, h,\text{ we get: } \\ \dfrac{dr}{dt}=\dfrac{-28}{35} \\ {{r}^{3}}+{{\left( r-h \right)}^{2}} \\ \text{Differentiating above equation partially wrt }t\text{ we get:} \\  \text{3}{{\text{r}}^{2}}\dfrac{dr}{dt}+(\dfrac{dr}{dt}-\dfrac{dh}{dt})(r-h)=0 \\ \dfrac{dr}{dt}=\dfrac{2(r-h)h^\prime }{3{{r}^{2}}+2(r-h)} \\ \dfrac{dr}{dt}=\dfrac{-4}{25} \\ \text{Now product of }\dfrac{-28}{35}\times \dfrac{-25}{4}=5 $ 