# Mathematics Questions and Answers – Derivatives Application – Tangents and Normals

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This set of Mathematics Multiple Choice Questions & Answers (MCQs) focuses on “Derivatives Application – Tangents and Normals”.

1. Find the tangent to the curve y=3x2+x+4 at x=3.
a) 19
b) 1.9
c) 18
d) 16

Explanation: The slope of the tangent at x=3 is given by
$$\frac{dy}{dx}$$]x=3= 6x+1]x=3=18+1=19.

2. Find the slope of the tangent to the curve x=4 cos3⁡3θ and y=5 sin3⁡⁡3θ at θ=π/4.
a) –$$\frac{3}{4}$$
b) –$$\frac{1}{4}$$
c) $$\frac{5}{4}$$
d) –$$\frac{5}{4}$$

Explanation: Given that, x=4 cos3⁡3θ and y=5 sin3⁡3θ
$$\frac{dx}{dθ}$$=4(3)(3 cos2⁡3θ)(-sin⁡3θ)
$$\frac{dy}{dθ}$$=5(3)(3 sin2⁡3θ)(cos⁡3θ)
$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{5(3)(3 sin^2⁡3θ)(cos⁡3θ)}{4(3)(3 cos^23θ)(-sin⁡3θ)}$$
$$\frac{dy}{dx}$$=-$$\frac{5 tan⁡3θ}{4}$$
$$\frac{dy}{dx}$$]θ=π/4=-$$\frac{5}{4} tan\frac{⁡3 \pi}{4}$$=-$$\frac{5}{4} (-1)=\frac{5}{4}$$.

3. Find the equation of all the lines having slope 0 which are tangent to the curve y=6x2-7x.
a) $$\frac{24}{49}$$
b) –$$\frac{24}{49}$$
c) $$\frac{49}{24}$$
d) –$$\frac{49}{24}$$

Explanation: Given that, y=6x2-7x
$$\frac{dy}{dx}$$=12x-7
It is given that the slope of tangent is 0
∴12x-7=0
⇒x=$$\frac{7}{12}$$
∴ y]x=$$\frac{7}{12}$$=6($$\frac{7}{12}$$)2-7($$\frac{7}{12}$$)
6($$\frac{49}{144})-\frac{49}{12}=49(-\frac{1}{24})=-\frac{49}{24}$$
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4. Find the points on the curve y=3x4+2x3-1 at which the tangents is parallel to the x-axis.
a) (0,1) and $$(-\frac{1}{2},-\frac{15}{16})$$
b) (0,-1) and $$(-\frac{1}{2},-\frac{15}{16})$$
c) (0,-1) and $$(\frac{1}{2},-\frac{15}{16})$$
d) (0,1) and $$(\frac{1}{2},\frac{15}{16})$$

Explanation: Given that, y=3x4+2x3-1
Differentiating w.r.t x, we get
$$\frac{dy}{dx}$$=12x3+6x2
The tangent is parallel to x-axis, which implies that the slope $$\frac{dy}{dx}$$ is 0.
∴12x3+6x2=0
6x2 (2x+1)=0
⇒x=0,-$$\frac{1}{2}$$
If x=0
y=3(0)+2(0)-1=-1
If x=-$$\frac{1}{2}$$
y=3$$(-\frac{1}{2})^4+2(-\frac{1}{2})^3-1$$
y=$$\frac{3}{16}-\frac{2}{8}-1$$
y=-$$\frac{15}{16}$$
Hence, the points at which the tangent is parallel to the x-axis is (0,-1) and $$(-\frac{1}{2},-\frac{15}{16})$$.

5. Find the tangent to the curve y=7x3-2x2 at the point x=2.
a) 67
b) 76
c) 46
d) 64

Explanation: The slope of the tangent at x=2 is given by
$$\frac{dy}{dx}$$]x=2 = 21x2-4x]x=2=21(2)2-4(2)=84-8=76
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6. Find the equation of the normal to the curve x=12 cosec⁡θ and y=2 sec⁡θ at x=π/4 .
a) $$\frac{1}{6}$$
b) -6
c) 6
d) –$$\frac{1}{6}$$

Explanation: Given that, x=12 cosecθ and y=2 sec⁡θ
$$\frac{dx}{dθ}$$=-12 cosec θ cot⁡θ
$$\frac{dy}{dθ}$$=2 tan⁡θ sec⁡θ
$$\frac{dy}{dx}$$=$$\frac{dy}{dθ}.\frac{dθ}{dx}=\frac{2 \,tan⁡θ \,sec⁡θ}{-12 \,cosec θ \,cot⁡θ}$$=-$$\frac{1}{6} \frac{sin⁡θ}{cos^2⁡θ} × \frac{cos⁡θ}{sin^2⁡⁡θ}$$ = –$$\frac{cot⁡θ}{6}$$
$$\frac{dy}{dx}]_{x=\frac{\pi}{4}}$$=$$-\frac{\frac{cot\pi}{4}}{6}=-\frac{1}{6}$$
Hence, the slope of normal at θ=π/4 is
–$$\frac{1}{slope \,of \,tangent \,at \,θ=\pi/4}=-\frac{-1}{\frac{1}{6}}$$=6

7. Find the equation of the tangent of the tangent to the curve 2x2+3y2=3 at the point(3,4).
a) x+2y=11
b) x-2y=11
c) -x+2y=11
d) x-2y=-11

Explanation: Differentiating 2x2+3y2=3 w.r.t x, we get
4x+6y $$\frac{dy}{dx}$$=0
$$\frac{dy}{dx} = -\frac{2x}{3y}$$
$$\frac{dy}{dx}$$](3,4)=-$$\frac{2(3)}{3(4)}=-\frac{1}{2}$$
Therefore, the equation of the tangent at (3,4) is
y-y0=m(x-x0)
y-4=-$$\frac{1}{2}$$ (x-3)
2(y-4)=-x+3
2y-8=-x+3
x+2y=11

8. Find the point at which the tangent to the curve y=$$\sqrt{4x^2+1}$$-2 has its slope2.
a) ($$\frac{1}{\sqrt{12}}$$,-1) and ($$\frac{1}{\sqrt{12}}$$,-1)
b) (-$$\frac{1}{\sqrt{12}}$$,3) and (-$$\frac{1}{\sqrt{12}}$$,-1)
c) ($$\frac{1}{\sqrt{12}}$$,2) and (-$$\frac{1}{\sqrt{12}}$$,-2)
d) ($$\frac{1}{\sqrt{12}}$$,3) and ($$\frac{1}{\sqrt{12}}$$,-2)

Explanation: Given that, y=$$\sqrt{4x^2+1}$$-2
$$\frac{dy}{dx}$$=$$\frac{1}{2\sqrt{4x^2+1}} (16x)=\frac{16x}{2\sqrt{4x^2+1}}$$
Given that the slope is 2
∴2=$$\frac{16x}{2\sqrt{4x^2+1}}$$
$$\sqrt{4x^2+1}$$=4x
4x2+1=16x2
12x2=1
x=±$$\frac{1}{\sqrt{12}}$$
When x=+$$\frac{1}{\sqrt{12}}$$
y=$$\frac{16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{16}{2\sqrt{12} (2/\sqrt{3})}=2$$
When x=-$$\frac{1}{\sqrt{12}}$$ y=$$\frac{-16}{2\sqrt{12} (\sqrt{(1/3+1)})}=\frac{-16}{2√12 (2/√3))}=-2$$
Hence, the point on the curve is ($$\frac{1}{\sqrt{12}}$$,2) and (-$$\frac{1}{\sqrt{12}}$$,-2)

9. Find the tangent to the curve y=5x4-3x2+2x-1 at x=1.
a) 15
b) 14
c) 16
d) 17

Explanation: The slope of the tangent at x=1 is given by
$$\frac{dy}{dx}$$]x=1= 20x3-6x+2]x=1=20(1)3-6(1)+2=20-6+2=16

10. Find the slope of the normal to the curve y=4x2-14x+5 at x=5.
a) –$$\frac{1}{26}$$
b) $$\frac{1}{26}$$
c) 26
d) -26

Explanation: The slope of the tangent at x=5 is given by:
$$\frac{dy}{dx}$$=8x-14
$$\frac{dy}{dx}$$]x=5=8(5)-14=40-14=26
∴ The slope of the normal to the curve is
$$-\frac{1}{slope\, of \,the \,tangent \,at \,x=5}=-\frac{1}{26}$$.

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