Years
>
Year 11>
Derivative Curve Sketching>
Log and Exponential Tangents and Normals>
Log and Exponential Tangents and NormalsLog and Exponential Tangents and Normals
Try VividMath Premium to unlock full access
Time limit: 0
Quiz summary
0 of 4 questions completed
Questions:
- 1
- 2
- 3
- 4
Information
–
You have already completed the quiz before. Hence you can not start it again.
Quiz is loading...
You must sign in or sign up to start the quiz.
You have to finish following quiz, to start this quiz:
Loading...
- 1
- 2
- 3
- 4
- Answered
- Review
-
Question 1 of 4
1. Question
Find the equation of the tangent to the curve `y=ln(5x+5)` at the point `x=3`.
Hint
Help VideoCorrect
Well Done!
Incorrect
Derivative of `y=ln f(x)`
$$y’=\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$$Point Gradient Formula
`y-``y_1``=``m``(x-``x_1``)`First, substitute `x=3` to the function to get the value of `y`.`y` `=` `ln(5``x``+5)` `=` `ln(5`(`3``)+5)` Substitute known values `=` `ln20` To find the slope, find the derivative of the function then substitute the value of `x``x` `=` `3` `f(x)` `=` `5x+5` `m` `=` `y’` `=` $$\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$$ `y’=(f'(x))/(f(x))` `=` $$\frac{\color{#9a00c7}{f'(5x+5)}}{\color{#D800AD}{5x+5}}$$ Substitute known values `=` $$\frac{5}{5(\color{#e65021}{x})+5}$$ Differentiate `5x+5` `=` $$\frac{5}{5(\color{#e65021}{3})+5}$$ Substitute known values `=` `5/(20)` `=` `1/4` Simplify Next, substitute the components to the point gradient formula.`x_1` `=` `3` `y_1` `=` `ln20` `m` `=` `1/4` `y-``y_1` `=` `m``(x-``x_1``)` `y-``ln20` `=` `1/4``(x-``3``)` Substitute known values `(y-ln20)``times4` `=` `(1/4(x-3))``times4` Multiply both sides by `4` `4y-4ln20` `=` `x-3` `4y-4ln20-x+3` `=` `0` Move all terms to the left side `4y-4ln20-x+3=0` -
Question 2 of 4
2. Question
Given the function `y=e^(2x)`, find the equation of the tangent line to the point `(2,e^4)`Hint
Help VideoCorrect
Fantastic!
Incorrect
Product Rule with Base “e”
$$y’=\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$Point Gradient Formula
`y-``y_1``=``m``(x-``x_1``)`To find the slope, find the derivative of the function then substitute the value of `x``x` `=` `2` `m` `=` `y’` `=` $$\color{#9a00c7}{f'(x)}\cdot e^{\color{#D800AD}{f(x)}}$$ `y’=f'(x) e^(f(x))` `=` $$\color{#9a00c7}{f'(2x)}\cdot e^{\color{#D800AD}{2x}}$$ Substitute known values `=` $$2\cdot e^{2\color{#e65021}{x}}$$ Differentiate `2x` `=` $$2e^{2(\color{#e65021}{2})}$$ Substitute known values `=` `2e^4` Next, substitute the components to the point gradient formula.`x_1` `=` `2` `y_1` `=` `e^4` `m` `=` `2e^4` `y-``y_1` `=` `m``(x-``x_1``)` `y-``e^4` `=` `2e^4``(x-``2``)` Substitute known values `y-e^4` `=` `2xe^4-4e^4` Distribute `y-e^4` `+e^4` `=` `2xe^4-4e^4` `+e^4` Add `e^4` to both sides `y` `=` `2xe^4-3e^4` `y=2xe^4-3e^4` -
Question 3 of 4
3. Question
Find the slope (`m`) and the angle of inclination (`theta`) at:`x=4`Round your answer to two decimal places-
`m=` (54.98)`theta=` (89)`°`
Hint
Help VideoCorrect
Nice Job!
Incorrect
Slope of an Exponential Function
`y’=e^x`To find the slope, find the derivative of the function then substitute the value of `x``m` `=` `y’` `=` `e^x` `y’=e^x` `=` `e^4` Substitute known values `=` `54.98` Compute using calculator Next, solve for the angle of inclination using `tantheta=``m`, where `theta` is the angle of inclination.`tantheta` `=` `m` `theta` `=` `tan^-1``(m)` Reverse the function to get `theta` `theta` `=` `tan^-1``(54.98)` Substitute known values `theta` `=` `89°` Compute using calculator `m=54.98``theta=89°` -
-
Question 4 of 4
4. Question
Find the equation of the normal line that intersects the curve `y=ln sqrtx` at the point `y=-1`.
Hint
Help VideoCorrect
Excellent!
Incorrect
Derivative of `y=ln f(x)`
$$y’=\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$$Point Gradient Formula
`y-``y_1``=``m``(x-``x_1``)`First, substitute `y=-1` to the function to get the value of `x`.`y` `=` `ln sqrtx` `-1` `=` `ln sqrtx` Substitute known values `-1` `=` $$\ln x^{\color{#CC0000}{\frac{1}{2}}}$$ Change the surd into an exponent $$\color{#CC0000}{e}^{-1}$$ `=` $$\color{#CC0000}{e}^{\ln x^{\frac{1}{2}}}$$ Insert `e` as base to both sides `e^(-1)` `=` $$x^{\frac{1}{2}}$$ `e^(ln x)=x` $$(e^{-1})^\color{#CC0000}{2}$$ `=` $$(x^{\frac{1}{2}})^\color{#CC0000}{2}$$ Raise both sides by `2` `e^(-2)` `=` `x` `x` `=` `1/(e^2)` Reciprocate `e^(-2)` To find the tangent, find the derivative of the function then substitute the value of `x``x` `=` `1/(e^2)` `f(x)` `=` `x^(1/2)` `y’` `=` $$\frac{\color{#9a00c7}{f'(x)}}{\color{#D800AD}{f(x)}}$$ `=` $$\frac{\color{#9a00c7}{f'(x^{\frac{1}{2}})}}{\color{#D800AD}{x^{\frac{1}{2}}}}$$ Substitute known values `=` $$\frac{\frac{1}{2}x^{-\frac{1}{2}}}{x^{\frac{1}{2}}}$$ Differentiate `x^(1/2)` `=` `1/(x^(1/2)) times 1/(2x^(1/2))` Simplify `=` $$\frac{1}{2\color{#e65021}{x}}$$ `=` $$\frac{1}{2\color{#e65021}{\frac{1}{e^2}}}$$ Substitute known values `=` `1/(2/(e^2))` `=` `(e^2)/2` Reciprocate `2/(e^2)` To find the normal line, get the negative reciprocal of the tangent, which is `(e^2)/2``(e^2)/2` `=` `-2/(e^2)` Get the negative reciprocal Next, substitute the components to the point gradient formula.`x_1` `=` `1/(e^2)` `y_1` `=` `-1` `m` `=` `-2/(e^2)` `y-``y_1` `=` `m``(x-``x_1``)` `y-``-1` `=` `-2/(e^2)``(x-``1/(e^2)``)` Substitute known values `y+1` `=` `-2/(e^2)x+2/(e^4)` Distribute `y+1` `-1` `=` `-2/(e^2)x+2/(e^4)` `-1` Subtract `1` from both sides `y` `=` `-2/(e^2)x+2/(e^4)-1` `y=-2/(e^2)x+2/(e^4)-1`