Difference between revisions of "009A Sample Midterm 1, Problem 3"

Revision as of 13:29, 16 February 2017

Let $y={\sqrt {3x-5}}.$

a) Use the definition of the derivative to compute

{\frac {dy}{dx}}

for

y={\sqrt {3x-5}}.

b) Find the equation of the tangent line to

y={\sqrt {3x-5}}

at

(2,1).

Foundations:
1. Limit Definition of Derivative
2. Tangent line equation

Solution:

(a)

Step 1:
Let $f(x)={\sqrt {3x-5}}.$
Using the limit definition of the derivative, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {f(x+h)-f(x)}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3(x+h)-5}}-{\sqrt {3x-5}}}{h}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {{\sqrt {3x+3h-5}}-{\sqrt {3x-5}}}{h}}.}\end{array}}$

Step 2:
Now, we multiply the numerator and denominator by the conjugate of the numerator.
Hence, we have
${\begin{array}{rcl}\displaystyle {f'(x)}&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {({\sqrt {3x+3h-5}}-{\sqrt {3x-5}})}{h}}{\frac {({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}{({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {(3x+3h-5)-(3x-5)}{h({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3h}{h({\sqrt {3x+3h-5}}+{\sqrt {3x-5}})}}}\\&&\\&=&\displaystyle {\lim _{h\rightarrow 0}{\frac {3}{{\sqrt {3x+3h-5}}+{\sqrt {3x-5}}}}}\\&&\\&=&\displaystyle {\frac {3}{{\sqrt {3x-5}}+{\sqrt {3x-5}}}}\\&&\\&=&\displaystyle {{\frac {3}{2{\sqrt {3x-5}}}}.}\end{array}}$

(b)

Step 1:
We start by finding the slope of the tangent line to $f(x)={\sqrt {3x-5}}$ at $(2,1).$
Using the derivative calculated in part (a), the slope is
${\begin{array}{rcl}\displaystyle {m}&=&\displaystyle {f'(2)}\\&&\\&=&\displaystyle {\frac {3}{2{\sqrt {3(2)-5}}}}\\&&\\&=&\displaystyle {{\frac {3}{2}}.}\end{array}}$

Step 2:
Now, the tangent line to $f(x)={\sqrt {3x-5}}$ at $(2,1)$
has slope $m={\frac {3}{2}}$ and passes through the point $(2,1).$
Hence, the equation of this line is
$y={\frac {3}{2}}(x-2)+1.$

Final Answer:
(a) ${\frac {3}{2{\sqrt {3x-5}}}}$
(b) $y={\frac {3}{2}}(x-2)+1$

@@ Line 18: / Line 18: @@
 {| class="mw-collapsible mw-collapsed" style = "text-align:left;"
 !Step 1: &nbsp;
+|-
+|Let <math>f(x)=\sqrt{3x-5}.</math>
 |-
 |Using the limit definition of the derivative, we have
@@ Line 57: / Line 59: @@
 !Step 1: &nbsp;
 |-
-|
+|We start by finding the slope of the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1).</math>
 |-
-|
+|Using the derivative calculated in part (a), the slope is
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>\begin{array}{rcl}
-|-
+\displaystyle{m} & = & \displaystyle{f'(2)}\\
-|
+&&\\
+& = & \displaystyle{\frac{3}{2\sqrt{3(2)-5}}}\\
+&&\\
+& = & \displaystyle{\frac{3}{2}.}
+\end{array}</math>
 |}
@@ Line 69: / Line 75: @@
 !Step 2: &nbsp;
 |-
-|
+|Now, the tangent line to <math>f(x)=\sqrt{3x-5}</math> at <math>(2,1)</math>
 |-
-|
+|has slope <math>m=\frac{3}{2}</math> and passes through the point <math>(2,1).</math>
 |-
-|
+|Hence, the equation of this line is
 |-
-|
+|&nbsp; &nbsp; &nbsp; &nbsp; <math>y=\frac{3}{2}(x-2)+1.</math>
 |}
@@ Line 81: / Line 87: @@
 !Final Answer: &nbsp;
 |-
-|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>f'(x)=\frac{3}{2\sqrt{3x-5}}</math>
+|&nbsp; &nbsp; '''(a)''' &nbsp; &nbsp; <math>\frac{3}{2\sqrt{3x-5}}</math>
 |-
-|'''(b)'''
+|&nbsp; &nbsp; '''(b)''' &nbsp; &nbsp; <math>y=\frac{3}{2}(x-2)+1</math>
 |}
 [[009A_Sample_Midterm_1|'''<u>Return to Sample Exam</u>''']]