Solveeit Logo
Login

Question

Question: The slope of the tangent to the curve \({\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right...

The slope of the tangent to the curve (y−x5)2=x(1+x2)2{\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2} at the point (1,3)\left( {1,3} \right) is

Explanation

Solution

Tangent is a line that touches the curve at only one point.
Use differentiation to find the slope of the tangent to the curve at the given point.
The given function is an implicit function, in which xx is dependent and yy is an independent variable, given in terms of both the variables xx and yy.
For example: x2+6xy+2y2=0{x^2} + 6xy + 2{y^2} = 0 is an explicit function.
Thus, after differentiating the given function except for the derivative dydx\dfrac{{dy}}{{dx}} , try to bring all the other terms on the same side of the equation.
Remember that the derivative dydx\dfrac{{dy}}{{dx}} is the required slope of the tangent.

Complete step-by-step answer:
Given that: (y−x5)2=x(1+x2)2{\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2}
On differentiating both sides with respect to xx .
Using differential: d(x)ndx=nxn−1\dfrac{{d{{\left( x \right)}^n}}}{{dx}} = n{x^{n - 1}}
2(y−x5)(dydx−5x4)=(1+x2)2+2x(1+x2)(2x)2\left( {y - {x^5}} \right)\left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = {\left( {1 + {x^2}} \right)^2} + 2x\left( {1 + {x^2}} \right)\left( {2x} \right)
⇒(dydx−5x4)=(1+x2)2+4x2(1+x2)2(y−x5)\Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{{{\left( {1 + {x^2}} \right)}^2} + 4{x^2}\left( {1 + {x^2}} \right)}}{{2\left( {y - {x^5}} \right)}}
⇒(dydx−5x4)=(1+x2)[(1+x2)+4x2]2(y−x5)\Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{\left( {1 + {x^2}} \right)\left[ {\left( {1 + {x^2}} \right) + 4{x^2}} \right]}}{{2\left( {y - {x^5}} \right)}}
⇒(dydx−5x4)=(1+x2)(1+5x2)2(y−x5)\Rightarrow \left( {\dfrac{{dy}}{{dx}} - 5{x^4}} \right) = \dfrac{{\left( {1 + {x^2}} \right)\left( {1 + 5{x^2}} \right)}}{{2\left( {y - {x^5}} \right)}}
⇒dydx=(1+x2)(1+5x2)2(y−x5)+5x4\Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{\left( {1 + {x^2}} \right)\left( {1 + 5{x^2}} \right)}}{{2\left( {y - {x^5}} \right)}} + 5{x^4}
Find the slope of the tangent:
The slope of the tangent to the curve y=f(x)y = f\left( x \right) at the point (x0,y0)\left( {{x_0},{y_0}} \right) is given by dydx](x0,y0){\left. {\dfrac{{dy}}{{dx}}} \right]_{\left( {{x_0},{y_0}} \right)}}
dydx](1,3)=(1+12)(1+5(1)2)2(3−15)+5(1)4{\left. {\dfrac{{dy}}{{dx}}} \right]_{\left( {1,3} \right)}} = \dfrac{{\left( {1 + {1^2}} \right)\left( {1 + 5{{\left( 1 \right)}^2}} \right)}}{{2\left( {3 - {1^5}} \right)}} + 5{\left( 1 \right)^4}
⇒2×62×2+5 ⇒3+5 ⇒8  \Rightarrow \dfrac{{2 \times 6}}{{2 \times 2}} + 5 \\\ \Rightarrow 3 + 5 \\\ \Rightarrow 8 \\\

The slope of the tangent to the curve (y−x5)2=x(1+x2)2{\left( {y - {x^5}} \right)^2} = x{\left( {1 + {x^2}} \right)^2} at the point (1,3)\left( {1,3} \right) is 8.

Note: In a similar question equation of the tangent can be asked further. Use the equation of the straight line passing through a given point (x0,y0)\left( {\mathop x\nolimits_0 ,\mathop y\nolimits_0 } \right) having finite slope mm:
y−y0=m(x−x0)y - \mathop y\nolimits_0 = m\left( {x - \mathop x\nolimits_0 } \right)
Another entity, normal, is associated with the tangent.
The normal line to the curve at a given point is perpendicular to the tangent at that point.
Thus the slope of the normal line =−1slope of tangent = - \dfrac{1}{{{\text{slope of tangent}}}}
We know, the slope of the tangent line to the curve y=f(x)y = f\left( x \right) is given by
The slope of the tangent line =dydx = \dfrac{{dy}}{{dx}} or f′(x)f'\left( x \right)
Thus, the slope of the normal line to the curve y=f(x)y = f\left( x \right) is given by
The slope of the normal line =−1dydx or −1f′(x) = - \dfrac{1}{{\dfrac{{dy}}{{dx}}}}{\text{ or }} - \dfrac{1}{{f'\left( x \right)}}