- Difficulty
- Introductory
- Suggested time
- 3 min
- Marks
- 2
Implicit differentiation and inverse trigonometric functions practice question 1
The point lies on the circle . Use implicit differentiation to find the slope of the tangent line there.
You are not given as a formula in . How can you differentiate the equation of the circle as it stands?
Differentiate both sides with respect to , remembering the factor on the term, then substitute the point.
A fraction, or a decimal to two decimal places.
Theory UBC Math 100 §4.5CLP-1 §2.11Key results
Worked solution
Strategy
Differentiate the relation as it stands, solve for y′, and substitute the coordinates of the point. The result agrees with geometry: the tangent to a circle is perpendicular to the radius.
Working
Implicit differentiation; the constant 25 differentiates to 0.
The radius to (3, 4) has slope 4/3; the tangent is perpendicular, slope −3/4.
Answer
The answer is negative three quarters.
Check
The circle passes through (3.01, 3.9925), where y has dropped by about 0.0075 for a step of 0.01 in x, a slope of about −0.75.
Where this goes wrong
Differentiating y² as 2y, leaving out y′. Then 2x + 2y = 0 has no derivative in it to solve for, and it is false at (3, 4). The chain-rule factor y′ is what makes the differentiated equation about slopes.
Leans on
The chain rule, from The chain rule and logarithms
Try a similar question
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