If the normal to the curve y = f(x) at the point (3, 4) makes an angle 3π/4 with the positive x-axis

Q: Wait, why is the slope of the normal $\tan(3\pi/4)$? Can't we just use the derivative directly?

Great question! The derivative $f'(3)$ gives the slope of the *tangent* line, not the normal. The normal is a different line entirely, perpendicular to the tangent. We use the angle of inclination of the normal to find its slope first, then use the perpendicularity condition to bridge the gap to the tangent's slope.

Q: Why is the product of the slopes $-1$? Is this always true?

Yes, for any two non-vertical lines that are perpendicular to each other in a Cartesian plane, the product of their slopes is always $-1$. This is a fundamental property derived from the geometry of right-angled triangles and rotation.

Q: I got $-1$ instead of $1$. Where did I go wrong?

It sounds like you might have forgotten that the product of the slopes is $-1$. If $m_n = -1$, then $m_t \cdot (-1) = -1$. Dividing both sides by $-1$ gives $m_t = 1$. Check your signs again—it is very easy to lose a negative sign in these calculations!

Question

If the normal to the curve $y = f(x)$ at the point $(3, 4)$ makes an angle $\frac{3\pi}{4}$ with the positive $x$-axis, then $f'(3) =$

Options:

(A) $-1$

(B) $-\frac{3}{4}$

(C) $\frac{4}{3}$

(D) $1$

Sigma Percentile · Accepted Answer

### Analyzing the Geometry of the Curve

Imagine you are standing on a smooth, elegant curve defined by the function $y = f(x)$. You are at the point $P(3, 4)$.

This coordinate is the heart of our problem. We are looking for the derivative $f'(3)$, which represents the slope of the tangent line at this exact location.

### The Normal Line

We draw a normal line at $P$. A normal is, by definition, the line perpendicular to the tangent at the point of contact.

The problem states that this line makes an angle of $	heta = \frac{3\pi}{4}$ with the positive $x$-axis. This angle serves as our primary anchor for the calculation.

### The Slope of the Normal

The slope of any line is the tangent of its angle of inclination. Therefore, the slope of the normal, $m_n$, is given by:

$$m_n = 	an\left(\frac{3\pi}{4}ight)$$

Since $\frac{3\pi}{4}$ corresponds to $135^{\circ}$ in the second quadrant, the tangent value is negative. Thus, we find:

$$m_n = -1$$

### The Tangent Connection

We know the tangent line is perpendicular to the normal line. In coordinate geometry, if two lines are perpendicular, the product of their slopes must be $-1$.

Let $m_t$ be the slope of the tangent. The relationship is defined as:

$$m_t \cdot m_n = -1$$

Substituting our known value for $m_n$, we get:

$$m_t \cdot (-1) = -1$$

### The Derivative

There is a fundamental bridge between geometry and calculus: the derivative of a function at a point is exactly the slope of the tangent line at that point.

Consequently, we identify that $m_t = f'(3)$. Substituting this into our previous equation, we have:

$$f'(3) \cdot (-1) = -1$$

### The Final Result

To find $f'(3)$, we divide both sides by $-1$. The negatives cancel out, leading us to the final conclusion:

**$f'(3) = 1$**

This is an elegant result that highlights the relationship between lines and curves. Always remember: when in doubt, visualize the geometry!

Animated Solution for Mathematics - Differentiation: If the normal to the curve $y = f (x)$ at the point $(3, 4)$ makes an angle $\frac{3 π}{4}$ with the positive $x$ -axis, then $f^{'} (3) =$

Select Answer:

Visualized Solution

The Curve and the Point

The Normal Line

Angle of Inclination

Slope Formula

Substituting the Angle

Calculating $m_{n}$

The Tangent Line

Perpendicular Slopes

Derivative is the Tangent Slope

Setting up the Equation

Solving for $f^{'} (3)$

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Geometry of the Curve

The Normal Line

The Slope of the Normal

The Tangent Connection

The Derivative

The Final Result

Similar Questions

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If the normal to the curve y=f(x) at the point (3,4) makes an angle 43π​ with the positive x-axis, then f′(3)=

Select Answer:

Visualized Solution

The Curve and the Point

The Normal Line

Angle of Inclination

Slope Formula

Substituting the Angle

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } mn​

The Tangent Line

Perpendicular Slopes

Derivative is the Tangent Slope

Setting up the Equation

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } f′(3)

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Geometry of the Curve

The Normal Line

The Slope of the Normal

The Tangent Connection

The Derivative

The Final Result

Similar Questions

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Animated Solution for Mathematics - Differentiation: If the normal to the curve $y = f (x)$ at the point $(3, 4)$ makes an angle $\frac{3 π}{4}$ with the positive $x$ -axis, then $f^{'} (3) =$

Calculating $m_{n}$

Solving for $f^{'} (3)$

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