Exercise 7.2 - Chapter 7 Applications of Differential Calculus 12th Maths Guide Samacheer Kalvi Solutions - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Ex $7.2$
Question 1.
Find the slope of the tangent to the curves at the respective given points.
(i) $y=x^4+2 x^2-x$ at $x=1$
(ii) $x=a \cos ^3 t, \sin ^3 \mathrm{tat} t=\frac{\pi}{2}$
Solution:
(i) $y=x^4+2 x^2-x$ at $x=1$
The slope of the tangent at $(x, y)$ is the value of $\frac{d y}{d x}$ at $\left(x_1, y_1\right)$
$
\begin{gathered}
\frac{d y}{d x}=4 x^3+4 x-1 \\
\frac{d y}{d x}(\text { at } x=1)=4+4-1=7
\end{gathered}
$
(i.e) slope of the tangent $=7$
(ii) $x=a \cos ^3 t, y=b \sin ^3 t$
$
\begin{aligned}
\frac{d x}{d t} & =a\left[3 \cos ^2 t\right](-\sin t) \\
\frac{d y}{d t} & =b\left[3 \sin ^2 t\right](\cos t) \\
\frac{d y}{d x} & =\frac{3 b \sin ^2 t \cos t}{-3 a \cos ^2 t \sin t}=\frac{-b}{a} \frac{\sin t}{\cos t} \\
\frac{d y}{d x}\left(\text { at } t=\frac{\pi}{2}\right) & =\frac{-b}{a} \frac{(1)}{0}=\infty
\end{aligned}
$
Slope of the tangent $=\infty$
Question 2.
Find the point on the curve $y=x^2-5 x+4$ at which the tangent is parallel to the line $3 x+y=7$.
Solution:
Let $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be the required point.

$
\begin{aligned}
& y=x^2-5 x+4 \\
& \frac{d y}{d x}=2 x-5 \\
& \frac{d y}{d x} \text { at }\left(x_1, y_1\right)=2 x_1-5=m_1=\text { slope of the tangent } \\
& 3 x+y=7 \\
& 3+\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-3=m_2
\end{aligned}
$
Given tangent is similarly to the line $\Rightarrow \mathrm{m}_1=\mathrm{m}_2$
$
\begin{aligned}
& \Rightarrow 2 \mathrm{x}_1-5=-3 \\
& \Rightarrow 2 \mathrm{x}_1=-3+5=2 \\
& \Rightarrow \mathrm{x}_1=1
\end{aligned}
$
Substituting $x_1=1$ in the curve.
$
\mathrm{y}_1=1-5+4=0
$
So the required point is $(1,0)$

Question 3.
Find the points on the curves $=x^3-6 x^2+x+3$ where the normal is parallel to the line $x+y=1729$.

Solution:
Let $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be the required point
$
y=x^3-6 x^2+x+3
$
$\begin{aligned} \frac{d y}{d x} & =3 x^2-12 x+1 \\ \frac{d y}{d x} \text { at }\left(x_1, y_1\right) & =3 x_1^2-12 x_1+1=m=\text { slope of the tangent }\end{aligned}$
$\therefore$ Slope of the normal $=\frac{-1}{m}=\frac{-1}{3 x_1^2-12 x_1+1}=m_1$ (say)
To find Slope of the line $x+y=1729$
$
1+\frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-1=m_2
$
Given that normal is parallel to the line. $m_1=m_2$
(i.e) $\frac{-1}{3 x_1^2-12 x_1+1}=-1$
$
\begin{aligned}
3 x_1^2-12 x_1+1=1 \Rightarrow 3 x_1{ }^2-12 x_1 & =0 \\
3 x_1\left(x_1-4\right) & =0 \\
x_1 & =0 \text { (or } 4
\end{aligned}
$
Substituting $x_1$ values in the curve
when $\mathrm{x}_1=4, \mathrm{y}_1=3$
when $\mathrm{x}_1=4, \mathrm{y}_1=4^3-6(4)^2+4+3=64-96+4+3=-25$
So the required points are $(0,3)$ and $(4,-25)$

Question 4.
Find the points on the curve $y^2-4 x y=x^2+5$ for which the tangent is horizontal.

Solution:
Let the required point be $\left(x_1, y_1\right) y^2-4 x y=x^2+5$
Slope of tangent at $\left(x_1, y_1\right)$ is $\frac{d y}{d x}$ at $\left(x_1, y_1\right)$ differentiating the curve w.r.to $x$
$
\begin{aligned}
2 y \frac{d y}{d x}-4\left[x \frac{d y}{d x}+y\right] & =2 x \\
\frac{d y}{d x}(2 y-4 x) & =2 x+4 y \\
\frac{d y}{d x} & =\frac{2 x+4 y}{2 y-4 x}=\frac{2(x+2 y)}{2(y-2 x)}=\frac{x+2 y}{y-2 x}
\end{aligned}
$
Given that tangent is horizontal (i.e) tangent is parallel to $\mathrm{x}$ - axis
$\Rightarrow$ Equation of tangent will be of the form $y=c$
$
\Rightarrow \quad \frac{d y}{d x}=0
$
from (1) and (2) we get
$
\frac{x+2 y}{y-2 x}=0 \Rightarrow x+2 y=0 \Rightarrow x=-2 y
$
Substituting $x=-2 y$ in the equation of the curve we get
$
\begin{aligned}
& \mathrm{y}^2-4(-2 \mathrm{y})(\mathrm{y})=4 \mathrm{y}^2+5 \\
& \Rightarrow \mathrm{y}^2+8 \mathrm{y}^2-4 \mathrm{y}^2=5 \\
& \Rightarrow 5 \mathrm{y}^2=5 \Rightarrow \mathrm{y}^2=1 \\
& \Rightarrow \mathrm{y}=\pm 1
\end{aligned}
$
when $\mathrm{y}=1, \mathrm{x}=-2$ and when $\mathrm{y}=-1, \mathrm{x}=2$ So the points are $(2,-1),(-2,1)$

Question 5.
Find the tangent and normal to the following curves at the given points on the curve.
(i) $y=x^2-x^4$ at $(1,0)$
(ii) $y=x^4+2 e^x$ at $(0,2)$
(iii) $y=x \sin x$ at $\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$
(iv) $x=\cos t, y=2 \sin ^2 t$ at $t=\frac{\pi}{3}$
Solution:

(i) $y=x^2-x^4$ $\begin{aligned} \frac{d y}{d x} & =2 x-4 x^3 \\ \frac{d y}{d x} \text { at }(1,0) & =2-4=-2=m=\text { Slope of the tangent }\end{aligned}$
Now $\mathrm{m}=-2,\left(\mathrm{x}_1, \mathrm{y}_1\right)=(1,0)$
So equation of the tangent is $\mathrm{y}-\mathrm{y}_1=\mathrm{m}\left(\mathrm{x}-\mathrm{x}_1\right)$
(i. e) $y-0=-2(x-1)$
$\mathrm{y}=-2 \mathrm{x}+2$
$2 x+y=2$
Slope of tangent $=m=-2$
$\therefore$ Slope of normal $=\frac{-1}{m}=\frac{1}{2} ;\left(x_1, y_1\right)=(1,0)$
So equation of normal is $y-y_1=\frac{-1}{m}\left(x-x_1\right)$
(i.e) $y-0=\frac{1}{2}(x-1)$
$
\begin{aligned}
2 y & =x-1 \\
x-2 y & =1
\end{aligned}
$

(ii) $y=x^4+2 e^x$ $\frac{d y}{d x}=4 x^3+2 e^x$ $\frac{d y}{d x}$ at $(0,2)=0+2(1)=2=m=$ slope of tangent
Now $m=2 ;\left(x_1, y_1\right)=(0,2)$
Equation of tangent is $y-2=2(x-0)$
$
\begin{aligned}
y-2 & =2 x \\
2 x-y & =-2
\end{aligned}
$
Slope of normal $=\frac{-1}{m}=\frac{-1}{2}$
Equation of normal is $y-2=\frac{-1}{2}(x-0)$
$
\begin{aligned}
& 2 y-4=-x \\
& x+2 y=4
\end{aligned}
$

(iii) $y=x \sin x$
$
\begin{aligned}
\frac{d y}{d x} & =x \cos x+\sin x \\
\frac{d y}{d x} \text { at }\left(\frac{\pi}{2}, \frac{\pi}{2}\right) & =0+1=1=m
\end{aligned}
$
Slope of the tangent $\left(x_1, y_1\right)=\left(\frac{\pi}{2}, \frac{\pi}{2}\right)$
Equation of tangent is $y-\frac{\pi}{2}=1\left(x-\frac{\pi}{2}\right)$
$
\begin{aligned}
x-y & =0 \\
\text { Slope of normal } & =\frac{-1}{m}=\frac{-1}{1}=-1
\end{aligned}
$
Equation of normal is $y-\frac{\pi}{2}=-1\left(x-\frac{\pi}{2}\right)$
$
\text { (i.e) } x+y=\pi
$

(iv) $x=\cos t, y=2 \sin ^2 t$
$
\begin{aligned}
x & =\cos t ; \frac{d x}{d t}=-\sin t \\
y & =2 \sin ^2 t ; \frac{d y}{d t}=2(2 \sin t \cos t)=4 \sin t \cos t \\
\frac{d y}{d x} & =\frac{d y}{d t} / \frac{d x}{d t}=\frac{4 \sin t \cos t}{-\sin t}=-4 \cos t
\end{aligned}
$
$\frac{d y}{d x}\left(\right.$ at $\left.t=\frac{\pi}{3}\right)=-4\left(\frac{1}{2}\right)=-2=m=$ slope of tangent
$
x_1=\cos \frac{\pi}{3}=\frac{1}{2}
$
$
y_1=2\left(\frac{\sqrt{3}}{2}\right)^2=2\left(\frac{3}{4}\right)=\frac{3}{2}
$
(i.e) $\left(x_1, y_1\right)=\left(\frac{1}{2}, \frac{3}{2}\right)$

Equation of tangent is $y-\frac{3}{2}=-2\left(x-\frac{1}{2}\right)$
$
\begin{aligned}
& 2 y-3=-2(2 x-1) \\
& 2 y-3=-4 x+2 \Rightarrow 4 x+2 y=5
\end{aligned}
$
Slope of tangent $=m=-2$
$
\therefore \text { Slope of normal }=\frac{-1}{m}=\frac{1}{2}
$
Equation of normal is $y-\frac{3}{2}=\frac{1}{2}\left(x-\frac{1}{2}\right)$
$
\begin{aligned}
2 y-3 & =\frac{2 x-1}{2} \\
4 y-6 & =2 x-1 \\
2 x-4 y & =-5
\end{aligned}
$

Question 6.
Find the equations of the tangents to the curve $y=1+x^3$ for which the tangent is orthogonal with the line $\mathrm{x}+12 \mathrm{y}=12$
Solution:
$
\begin{aligned}
& y=1+x^3 \\
& \frac{d y}{d x}=3 x^2 \\
& \qquad \frac{d y}{d x} \text { at }\left(x_1, y_1\right)=3 x_1^2=m_1=\text { Slope of the tangent. }
\end{aligned}
$
To find slope of $x+12 y=12$
$
1+12 \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=\frac{-1}{12}=m_2
$
Given that tangent is orthogonal to the lines.
$
\begin{aligned}
& \Rightarrow \text { Product of slopes }=-1 \\
& \Rightarrow\left(3 x_1^2\right)\left(-\frac{1}{12}\right)=-1 \\
& \Rightarrow\left(3 x_1^2\right)=12 \Rightarrow x_1^2=\frac{12}{3}=4 \Rightarrow x_1=\pm 2
\end{aligned}
$
Substituting $x_1$ values in the curve
when $x_1=2, y_1=9$; when $x_1=-2, y_1=-1$
So the points are $(2,9)$ and $(-2,-7)$
To find the equations of tangents:
Tangents are orthogonal to $\mathrm{x}+12 \mathrm{y}=12$
So equations of tangents will be of the form $12 \mathrm{x}-\mathrm{y}=\mathrm{k}$
The tangent passes through $(2,9) \Rightarrow 24-9=\mathrm{k} \Rightarrow \mathrm{k}=15$.
$\therefore$ Equation of tangent is $12 \mathrm{x}-\mathrm{y}=15$
The tangent passes through $(-2,-7) \Rightarrow 12(-2)+7=\mathrm{k} \Rightarrow-17$
So equation of tangent is $12 \mathrm{x}-\mathrm{y}=-17$

Question 7.
Find the equations of the tangents to the curve $y=\frac{x+1}{x-1}$ which are parallel to the line $\mathrm{x}+2 \mathrm{y}=6$.

Solution:
$
\begin{aligned}
& \qquad y=\frac{x+1}{x-1} \\
& \qquad \frac{d y}{d x}=\frac{(x-1)(1)-(x+1)(1)}{(x-1)^2}=\frac{-2}{(x-1)^2} \\
& \frac{d y}{d x} \text { at }\left(x_1, y_1\right)=\frac{-2}{\left(x_1-1\right)^2}=m_1=\text { Slope of tangent } \\
& \text { Tangent is parallel the line } \mathrm{x}+2 \mathrm{y}=6 \\
& \Rightarrow \text { Slope of tangent }=\text { Slope of line } \\
& \text { Slope of } x+2 y=6 \text { is }-\frac{1}{2} \\
& \text { So } \frac{-2}{\left(x_1-1\right)^2}=-\frac{1}{2} \Rightarrow\left(x_1-1\right)^2=4 \\
& \Rightarrow x_1^2+1-2 x_1-4=0 \\
& \Rightarrow x_1^2-2 x_1-3=0 \Rightarrow\left(x_1-3\right)\left(x_1+1\right)=0 \Rightarrow x_1=3 \text { or }-1
\end{aligned}
$
Substituting $x_1$ values in the curve $y=\frac{x+1}{x-1}$
when $\mathrm{x}_1=-1, \mathrm{y}_1=0$; when $\mathrm{x}_1=3, \mathrm{y}_1=2$
So the points are $(-1,0)$ and $(3,2)$. The tangents are parallel to $x+2 y=6$. So equation of tangents will be of the form $x+2 y=k$.
$\therefore$ Equation of tangent is $\mathrm{x}+2 \mathrm{y}=-1$
The tangent passes through $(-1,0) \Rightarrow-1=\mathrm{k}$
The tangent passes through $(3,2) \Rightarrow 3+4=\mathrm{k} \Rightarrow \mathrm{k}=7$
$\therefore$ Equation of tangent is $\mathrm{x}+2 \mathrm{y}=7$

Question 8.
Find the equation of tangent and normal to the curve given by $x=7 \cos t$ and $y=2 \sin t, t \in R$ at any point on the curve.
Solution:
$
\begin{aligned}
& x=7 \cos t \quad y=2 \sin t \\
& \frac{d x}{d t}=-7 \sin t \quad \frac{d y}{d t}=2 \cos t \\
& \frac{d y}{d x}\left[\text { at }{ }^{\prime} t^{\prime}\right]=\frac{2 \cos t}{-7 \sin t}=m=\text { slope of the tangent } \\
& \left(x_1, y_1\right)=(7 \cos t, 2 \sin t) \\
& \text { Equation of tangent is } y-2 \sin t=\frac{-2 \cos t}{7 \sin t}(x-7 \cos t) \\
& 7 y \sin t-14 \sin ^2 t=-2 x \cos t+14 \cos ^2 t \\
& 2 x \cos t+7 y \sin t=14\left(\cos ^2 t+\sin ^2 t\right)=14 \\
& \text {; } 2 x \cos t+7 y \sin t=14 \\
& \text { Slope of tangent }=m=\frac{-2 \cos t}{7 \sin t} \\
& \therefore \text { Slope of normal }=\frac{-1}{m}=\frac{7 \sin t}{2 \cos t} \\
& \left(x_1, y_1\right)=(7 \cos t, 2 \sin t) \\
&
\end{aligned}
$
So equation of normal is
$
y-2 \sin t=\frac{7 \sin t}{2 \cos t}(x-7 \cos t)
$
$(2 \cos t) y-4 \sin t \cos t=(7 \sin t) x-49 \sin t \cos t$
$
x(7 \sin t)-y(2 \cos t)=45 \sin t \cos t
$

Question 9.
Find the angle between the rectangular hyperbola $x y=2$ and the parabola $x^2+4 y=0$.
Solution:
Solving the given two equations. To find the point of intersection:
$
\begin{aligned}
x^2+4 y & =0 \\
4 y & =-x^2 \Rightarrow y=-\frac{x^2}{4}
\end{aligned}
$
Substituting $y=-\frac{x^2}{4}$ in $x y=2$ we get
$
x\left(\frac{-x^2}{4}\right)=2 \Rightarrow-x^3=8 \Rightarrow x=-2
$

$
\begin{aligned}
\text { when } x & =-2 \Rightarrow y=\frac{-(4)}{4}=-1 \\
\left(x_1, y_1\right) & =(-2,-1) \\
x y & =2
\end{aligned}
$
Differentiating w.r. to $x$
$
\begin{aligned}
x \frac{d y}{d x}+y & =0 \Rightarrow \frac{d y}{d x}=\frac{-y}{x} \\
\frac{d y}{d x} \text { at }(-2,-1) & =-\left(\frac{-1}{-2}\right)=\frac{-1}{2}=m_1 \\
x^2+4 y & =0
\end{aligned}
$
Differentiating w.r.to $x$
$
\begin{aligned}
2 x+4 \frac{d y}{d x} & =0 \Rightarrow \frac{d y}{d x}=\frac{-2 x}{4}=\frac{-x}{2} \\
\frac{d y}{d x} \text { at }(-2,-1) & =-\left(\frac{-2}{2}\right)=1=m_2
\end{aligned}
$
If $\theta$ is the angle between the curves then
$
\begin{aligned}
\tan \theta & =\left|\frac{m_1-m_2}{1+m_1 m_2}\right| \\
\text { (i.e) } \tan \theta & =\left|\frac{-\frac{1}{2}-1}{1+\left(\frac{-1}{2}\right)(+1)}\right|=\left|\frac{-\frac{3}{2}}{\frac{1}{2}}\right|=3 \Rightarrow \theta=\tan ^{-1}(3)
\end{aligned}
$

Question 10.
Show that the two curves $x^2-y^2=r^2$ and $x y=c^2$ where $c, r$ are constants, cut orthogonally.

Solution:
Let $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ be the point of intersection of the two curves
I Curve: $x^2-y^2=r^2$
Differentiating w.r.to $\mathrm{x}$

$
\begin{aligned}
2 x-2 y \frac{d y}{d x} & =0 \Rightarrow \frac{d y}{d x}=\frac{2 x}{2 y}=\frac{x}{y} \\
\frac{d y}{d x} \text { at }\left(x_1, y_1\right) & =\frac{x_1}{y_1}=m_1
\end{aligned}
$
II Curve: $x y=c^2$
Differentiating w.r.to $\mathrm{x}$
$
\begin{aligned}
x \frac{d y}{d x}+y & =0 \Rightarrow \frac{d y}{d x}=-\frac{y}{x} \\
\frac{d y}{d x} \text { at }\left(x_1, y_1\right) & =\frac{-y_1}{x_1}=m_2
\end{aligned}
$
Now $m_1 m_2=\left(\frac{x_1}{y_1}\right)\left(\frac{-y_1}{x_1}\right)=-1 \Rightarrow$ the two curves cut orthogonally.