Equation of a tangent
We can calculate the gradient of a tangent to a curve by differentiating.
In order to find the equation of a tangent, we:
- Differentiate the equation of the curve
- Substitute the \(x\) value into the differentiated equation to find the gradient
- Substitute the \(x\) value into the original equation of the curve to find the y-coordinate
- Substitute your point on the line and the gradient into \(y - b = m(x - a)\)
Example 1
Find the equation of the tangent to the curve \(y = \frac{1}{8}{x^3} - 3\sqrt x\) at the point where \(x = 4\).
Solution
First we need to get the equation of the curve into the form we can differentiate.
\(y = \frac{1}{8}{x^3} - 3{x^{\frac{1}{2}}}\)
\(\frac{{dy}}{{dx}} = \frac{3}{8}{x^2} - \frac{3}{2}{x^{ - \frac{1}{2}}}\)
\(\frac{{dy}}{{dx}} = \frac{{3{x^2}}}{8} - \frac{3}{{2\sqrt x }}\)
When \(x = 4\),
\(m = \frac{{3{{(4)}^2}}}{8} - \frac{3}{{2\sqrt 4 }}\)
\(= \frac{{3 \times 16}}{8} - \frac{3}{{2 \times 2}}\)
\(= 6 - \frac{3}{4}\)
\(= \frac{{21}}{4}\)
y-coordinate when \(x = 4\),
\(y = \frac{1}{8}{(4)^3} - 3{(4)^{\frac{1}{2}}}\)
\(= \frac{{64}}{8} - 3\sqrt 4\)
\(= 8 - 6\)
\(= 2\)
Equation of tangent when \(a = 4,\,b = 2,\,m = \frac{{21}}{4}\)
\(y - b = m(x - a)\)
\(y - 2 = \frac{{21}}{4}(x - 4)\)
\(4(y - 2) = 21(x - 4)\)
\(4y - 8 = 21x - 84\)
\(4y = 21x - 76\)
\(21x - 4y - 76 = 0\)
Therefore the equation of the tangent is \(21x - 4y - 76 = 0\)
You can also use this method to find the point of contact of a tangent to a curve when given the equation of the curve and the gradient of the tangent.
Example 2
Find the point of contact between the tangent and curve with equation \(y = 5{x^2} - 2x + 3\) when the gradient is \(\frac{4}{3}\).
Solution
First of all, we need to differentiate the equation of the curve to get an expression for the gradient.
\(\frac{{dy}}{{dx}} = 10x - 2\)
Now, we would usually 'sub-in' the \(x\) value to get the gradient, but this time we don't know what \(x\) is. We do know, however that the answer is \(\frac{4}{3}\), so we can set this equal to the differentiated expression and solve the equation to find \(x\).
\(10x - 2 = \frac{4}{3}\)
\(10x = \frac{{10}}{3}\)
\(x = \frac{1}{3}\)
Now we can find the y-coordinate by substituting this into the original equation.
\(y = 5{\left( {\frac{1}{3}} \right)^2} - 2\left( {\frac{1}{3}} \right) + 3\)
\(y = 5\left( {\frac{1}{9}} \right) - \frac{2}{3} + 3\)
\(y = \frac{5}{9} - \frac{6}{9} + \frac{{27}}{9}\)
\(y = \frac{{26}}{9}\)
Therefore the point of intersection is \(\left( {\frac{1}{3},\frac{{26}}{9}} \right)\)
