KCET · Maths · Straight Lines
The tangent to the curve \( y=x^{3}+1 \) at \( (1,2) \) makes an angle \( \theta \) with y axis, then the value of \( \tan \theta \) is
- A \( 03 \)
- B \( \frac{1}{3} \)
- C \( -\frac{1}{3} \)
- D \(-3 \)
Answer & Solution
Correct Answer
(C) \( -\frac{1}{3} \)
Step-by-step Solution
Detailed explanation
Given equation of curve
Here
\(
\theta=90^{\circ}+\phi . \text { Then }, \tan \theta=\tan \left(90^{\circ}+\phi\right)=-\cot \phi
\)
Now,
\(
\tan \theta=\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+1\right)=3 x^{2}
\)
At point \( (1,2) \), we have \( \left.\frac{d y}{d x}\right|_{(1,2)}=\left.3 x^{2}\right|_{(1,2)}=3 \)
Therefore, required value of \( \theta \) is
\(
-\cot \phi=-\frac{1}{3}
\)
In the above diagram, \( \theta \) is to be considered as the angle made by tangent with y-axis and not \( \theta^{\prime} \) [for e.g., when we say
Here
\(
\theta=90^{\circ}+\phi . \text { Then }, \tan \theta=\tan \left(90^{\circ}+\phi\right)=-\cot \phi
\)
Now,
\(
\tan \theta=\frac{d y}{d x}=\frac{d}{d x}\left(x^{3}+1\right)=3 x^{2}
\)
At point \( (1,2) \), we have \( \left.\frac{d y}{d x}\right|_{(1,2)}=\left.3 x^{2}\right|_{(1,2)}=3 \)
Therefore, required value of \( \theta \) is
\(
-\cot \phi=-\frac{1}{3}
\)
In the above diagram, \( \theta \) is to be considered as the angle made by tangent with y-axis and not \( \theta^{\prime} \) [for e.g., when we say
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